Physics



AP Physics 3: Forces—Dynamics Name __________________________

A. Newton’s Laws of Motion

1. Aristotle's view was that an object's natural state is rest and it takes a force (push or pull) to keep an object moving

2. Galileo's view was that an object's natural state was unchanged motion, either at rest or at a constant speed in a straight line (Law of Inertia)

3. Newton synthesized causes of motion into three laws

a. First Law (Galileo's law of Inertia): object remains at rest or uniform velocity in a straight line as long as no net force (Fnet) acts on it

b. Second Law: (Fnet = ma)

1. measured in newtons: 1 N = 1 kg•m/s

2. Fnet → and v →: v increases

Fnet ← and v →: v decreases

Fnet ↑ and v →: v turns in a circle

3. FΔt = mΔv

|Steps |Algebra |

|start with |F = ma |

|multiply both sides by Δt |FΔt = maΔt |

|substitute Δv/Δt for a |FΔt = m(Δv/Δt)Δt = mΔv |

a. mv is Newton's "quantity of motion"

b. now called momentum, p = mv (kg•m/s)

c. Third Law: action force on A generates an equal but opposite reaction force on B (FA = -FB)

d. four important concepts

1. force can act on contact (collision) or at a distance (gravity)

2. usually multiple forces act on an object ∴ the vector sum of all forces = Fnet

3. mass is measured in terms of Newton's laws

a. inertial mass = object's resistance to change in motion (first law)

b. gravitational mass = gravity's affect on an object (second law)

4. third law forces are equal and opposite, but don't cancel each other out because they act on different objects, which can cause either or both objects to accelerate.

B. Types of Forces

1. push or pull (Fp)

a. measured using a spring scale (force increases linearly as distance that a spring is stretched x increases)

1. spring force, Fs = kx

2. k is the spring constant

b. tension (Ft or T) can be used instead of Fp

2. weight (Fg or W) is the force of attraction between the object and the Earth—gravity, Fg = mg

a. g = 9.80 m/s2 (negative sign is not included)

b. directed down to the Earth’s center

3. normal force (Fn or N) is the force that the surface exerts on an object to support its weight

a. perpendicular away from the surface

b. not calculated in isolation, but is determined by other perpendicular forces so that ΣF⊥ = 0

4. friction (Ff) is parallel to surface and opposes motion

a. when moving: Ff ’ μkFn

(μk = kinetic coefficient of friction)

b. when stationary: Ff is part of ∑F|| = 0, but cannot exceed Ff ≤ μsFn (μs = static coefficient of friction)

5. free-body diagram

a. diagram shows all forces acting on the system

b. Fp/Ft: along direction of push or pull

c. Ff: opposes motion and is || to surface

d. Fg—toward Earth's center

e. Fn—⊥ to surface

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C. Force Problems—Dynamics

1. general set-up

|draw a free body diagram |

|resolve forces into || and ⊥ components to motion |

|assign positive directions |

|for perpendicular forces, up is positive |

|for parallel forces, direction of velocity is positive |

|two equations |

|ΣF⊥ = 0 |

|ΣF|| = ma |

|calculate d, v and t using kinematics |

2. horizontal surface

| Fp Fn |

|Fp-⊥ = Fpsinθ |

|θ Ff |

|Fp-|| = Fpcosθ |

| |

|Fn = Fg – Fp-⊥ Fg = mg |

|Ff = μFn |

|ΣF|| = Fp-|| – Ff = ma (m is everything that moves) |

3. incline (moving up)

| Fn |

|Fp |

| |

| |

|Ff θ |

|Fg-⊥ = Fgcosθ |

|θ Fg= mg |

|Fg-|| = Fgsinθ |

|Fn = Fg-⊥ |

|Ff = μFn |

|F|| = Fp – Ff – Fg-|| = ma (m is everything that moves) |

5. internal tension

|treat the system as one object |

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|Ff Ft-1 = Ft-2 Fp |

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|Ft-1 = Ft-2 (third law) ∴ cancel out |

|ΣF|| = Fp – Ff = ma (m is everything that moves) |

|a = (Fp – Ff)/(m1 + m2) |

|isolate one part |

|Fp – Ft-2 = m2a (a is for the whole system) |

|Ft-1 – Ff = m1a (a is for the whole system) |

6. pulleys

|treat the system as one object (m2 > m1) |

|T1 = T2 (third law) ∴ cancel out +a ↑ ↓ +a |

|ΣF|| = W2 – W1 = ma |

|a = (m2 – m1)g/(m2 + m1) T1 T2 |

|isolate one part m1 |

|m2 |

|W2 – T2 = m2a W1 W2 |

|T1 – W1 = m1a |

7. vertical acceleration

|Fn (platform) or Fp (rope, rocket) generates acceleration |

|Fn/p – Fg = ma |

|+a, apparent weight > normal |

|–a, apparent weight < normal |

|–a = g (weightless) |

D. Force Problems—Statics

1. one unknown

|draw a free body diagram |

|resolve forces into || and ⊥ components to motion |

|assign positive directions |

|two equations |

|ΣF⊥ = 0 |

|ΣF|| = 0 |

|calculate d, v and t using kinematics |

2. two unknowns

| θL θR |

|TL TR |

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| |

|Fg = mg |

|ΣFy = TLsin(180 – θL) + TRsin(θR) + Fgsin(-90) = 0 |

|ΣFx = TLcos(180 – θL) + TRcos(θR) + Fgcos(-90) = 0 |

|solve for TL in terms of TR in the second equation and then substitute |

|into the first equation |

|special case (two of three forces are ⊥) |

|θL |

|TL |

|Fg TL |

|θL TR |

|TR Fg = mg |

|sinθL = Fg/TL |

|tanθL = Fg/TR |

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A. Newton's Laws of Motion

Atwood Machine Lab

Time how long it takes the upper weight with additional mass mp to descend the measured distance, graph the data, calculate the acceleration using kinematics and Newton's law and compare the two accelerations.

a. Collect the following data.

|Descending Distance | |

|d (m) | |

|Exp. |mp |Fp |Time (s) |

| |(g) |(N) | |

| | | |trial 1 |trial 2 |trial 3 |Average |

|1 |15 |0.147 | | | | |

|2 |16 |0.157 | | | | |

|3 |17 |0.167 | | | | |

|4 |18 |0.176 | | | | |

|5 |19 |0.186 | | | | |

|6 |20 |0.196 | | | | |

|7 |21 |0.206 | | | | |

b. Calculate the acceleration using kinematics.

| |Formula |Calculation |

| | |1 |2 |3 |4 |

| |a (m/s2) |

d. Determine the y-intercept. This is Ff = ________

e. Calculate the net force, total mass and acceleration.

|Experiment |1 |2 |3 |4 |5 |6 |7 |

|Fnet |Fp - Ff| | | | | | |

|% Δ | | | | | | | |

|Average | |

Questions 1-11 Briefly explain your answer.

1. A book is lying at rest on a table because

(A) there are no forces acting on the book.

(B) the forces cancel each other out.

|B—Fg and Fn cancel each other out |

2. A hockey puck slides on ice at constant velocity. What is the direction of the net force on the puck?

(A) forward (B) backward (C) no net force

|C—Fnet = ma = m(0) = 0 |

3. You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward a short distance. What is the direction of the net force on the book?

(A) forward (B) backward (C) no net force

|B—Friction slows the book ∴ force is backward |

4. You kick a smooth flat stone out on a frozen pond. The stone slides, slows down and eventually stops. What is the direction of the net force on the stone?

(A) forward (B) backward (C) no net force

|B—Friction slows the stone ∴ force is backward |

5. Consider a cart on a horizontal frictionless table. Once the cart has been given a push and released, what will happen to the speed of the cart?

(A) slow down (B) constant (C) speed up

|B—Fnet = 0 ∴ no acceleration and the speed stays the same |

Questions 6-8 From rest, you step on the gas of a Ferrari, providing a force F for time t, resulting in final speed v and distance traveled d.

6. How much time would be needed to reach speed v using ½F?

(A) 4t (B) 2t (C) ½t (D) ¼t

|B—FΔt = mΔv, but mΔv is constant ∴ FΔt is constant |

|Ft = ½FΔt ∴ Δt = 2t |

7. How much time would be needed to reach a speed of 2v?

(A) 4t (B) 2t (C) ½t (D) ¼t

|B—FΔt = mΔv, double v doubles Δt |

8. How far would the car travel if t is doubled?

(A) 4d (B) 2d (C) ½d (D) ¼d

|A—FΔt = mΔv, double t doubles Δv, but Δv2 = 2ad ∴ double t quadruples d |

9. A force F acts on a mass M1, giving acceleration a1. The same force acts on a different mass M2, giving acceleration a2 = 2a1. If M1 and M2 are glued together and the same force F acts on this combination, what is the resulting acceleration?

(A) 3/2a1 (B) 1/2a1 (C) 2/3a1 (D) 1/3a1

|C—F = ma: a2 = 2a1 ∴ m2 = ½m1 and mtot = 3/2m1 |

|∴ a = 2/3a1 |

Questions 10-11 David Scott (Apollo 15) dropped a feather and hammer on the Moon from the same height and they reached the ground at the same time. Which object had a greater acceleration due to gravity?

(A) feather (B) hammer (C) tie

|C—Both fell at the same rate ∴ same acceleration. |

10. Which had the greater force of gravity?

(A) feather (B) hammer (C) tie

|B—Both have equal accelerations, but the hammer has greater mass (Fg = |

|ma) |

11. David Scott throws a ball on Earth. On the Moon, he throws the same ball with the same force. The acceleration of the ball on the Moon compared to Earth is

(A) more (B) less (C) the same

|C—F = ma: with the same F and m, a will be the same |

12. What is the direction of the net force for each situation?

|car is accelerating northward |northward |

|bowling ball rolling straight at constant speed |zero |

|thrown rock reaches its highest point |downward |

|rock resting on the ground |zero |

13. Determine the net force and acceleration on the 10-kg box for each situation.

| |Fnet = -30 N + 30 N = 0 N |

|30 N 30 N | |

| |Fnet = ma |

| |0 N = (10 kg)a ∴ a = 0 m/s2 |

| |Fnet = -30 N + 50 N = 20 N |

|30 N 50 N | |

| |Fnet = ma |

| |20 N = (10 kg)a ∴ a = 2 m/s2 |

| |Fnet = -30 N + 20 N = -10 N |

|30 N 20 N | |

| |Fnet = ma |

| |-10 N = (10 kg)a ∴ a = -1m/s2 |

14. State whether the pair of equal forces are third law forces or first law forces.

a. Force between two ice skaters pushing on each other.

|Third Law |

b. Tension in a cord equals the weight of a hanging mass.

|First Law |

15. Two 1-kg weights are suspended from a frictionless pulley (an Atwood machine). An additional 0.0500 kg is added to the upper weight and allowed to fall freely.

a. What is the sum of forces acting on the system?

|Fnet = mg = (0.0500 kg)(10 m/s2) = 0.500 N |

b. What is the total mass of all moving parts?

|mtotal = 1 kg + 1 kg + 0.05 kg + 2.05 kg |

c. What is the theoretical acceleration?

|Fnet = ma |

|0.500 N = (2.05 kg)a ∴ a = 0.244 m/s2 |

d. The system takes 2.90 s to move 1.00 m. What is the actual acceleration?

|d = ½at |

|1.00 m = ½a(2.90 s)2 ∴ a = 0.238 m/s2 |

e. What is the percent difference between theoretical and actual accelerations?

|% Δ = 100(0.244 – 0.238)/0.244 = 2 % |

B. Types of Forces

Static Friction Lab

Part 1: pull on the wood block with weights on top from the same initial spot near the top of the incline using the spring scale until the block just begins to move, record the greatest force reading on the spring scale (Fs), graph the data and use the slope to determine μs.

a. Collect the following data.

|Mass of wood block, M1 (kg) | |

|Weights, M2 (kg) |0 |0.10 |0.20 |0.30 |0.40 |0.50 |

|Fn |Fn = (M1 + M2)g| | | |

| |Fn |

c. Use the slope of the line to determine μs.

| |

Part 2: measure the angle of the incline, which is needed for the wood block to just begin moving (angle of repose).

d. Collect the following data

|Trial |1 |2 |3 |Average |

|θ | | | | |

|tanθ | |

e. Label all the forces acting on the block in terms of Fg.

Fn = Fgcosθ

Ff = Fgsinθ

θ Fg−⊥ = Fgcosθ

Fg

θ Fg-|| = Fgsinθ

f. Determine an expression for μ in terms of θ.

|μ = Ff/Fn |

|μ = Fgsinθ/Fgcosθ = tanθ |

g. Calculate the percent difference between μ (part c) and tanθ ( part d).

| |

h. Does placing the block on its side significantly change the angle of repose? Find out.

|Trial |1 |2 |3 |Average θ |

|Angle of repose | | | | |

|Affect on μ | |

Questions 16-22 Briefly explain your answer.

16. Consider two identical blocks; A resting on a horizontal surface and B resting on an incline. Which block has a greater normal force?

(A) NA > NB (B) NA < NB (C) NA = NB

|A—NA = Fg, but NB = Fgcosθ, which is less than Fg |

17. When you walk forward, what is the direction of the net force you exert on the ground?

(A) forward (B) backward (C) downward

|B—Newton's third law reaction force generates the forward force |

Questions 18-21 Below you see two cases: a physics student pushing (a) or pulling (b) a sled with a force F, which is applied at an angle θ.

18. Which is true of the normal force N on the sled?

(A) Na > Nb (B) Na < Nb (C) Na = Nb

|A—Na = mg + Fsinθ and Nb = mg – Fsinθ ∴ Na > Nb |

19. Which is true of the force of friction Ff on the sled?

(A) Ffa > Ffb (B) Ffa < Ffb (C) Ffa = Ffb

|A—Ffa = μNa and Ffb = μNb, but Na > Nb ∴ Ffa > Ffb |

20. Which is true of the force F need for constant speed?

(A) Fa > Fb (B) Fa < Fb (C) Fa = Fb

|A—since Fa = Ffa and Fb = Ffb, but Ffa > Ffb ∴ Fa > Fb |

21. Which is true of the acceleration if the force F is the same?

(A) aa > ab (B) aa < ab (C) aa = ab

|B—F – Ff = ma, since Ffa > Ffb, then less force is available for |

|acceleration ∴ aa < ab |

22. A box generates a maximum force of static friction equal to 50 N. A horizontal force of 30 N is directed to the right. What direction will the box move?

(A) right (B) left (C) the box doesn't move

|C—Friction can only resist motion, not cause it, since the horizontal |

|force < force of friction, then no motion |

23. A 100-N force Fp pulls on a 10 kg block at an angle of 30o from horizontal along a frictionless surface.

a. What is the perpendicular component of Fp?

|Fp-⊥ = Fpsin30 = (100 N)sin30 = 50 N |

b. What is the parallel component of Fp?

|Fp-|| = Fpcos30 = (100 N)cos30 = 87 N |

c. What is the acceleration of the block?

|ΣF|| = ma |

|87 N = (10 kg)a ∴ a = 8.7 m/s2 |

24. What is the force of gravity (weight) of a 75 kg person?

a. On earth (g = 9.8 m/s2)

|Fg = mgearth = (75 kg)(9.8 m/s2) = 735 N |

b. On the moon (g = 1.6 m/s2)

|Fg = mgmoon = (75 kg)(1.6 m/s2) = 120 N |

25. Determine the normal force on 10 kg block that is

a. resting on a horizontal surface.

|Fn = Fg = mg = (10 kg)(10 m/s2) = 100 N |

b. resting on a horizontal surface with a force of 25 N pressing down on the block.

|Fn = Fg + Fp = 100 N + 25 N = 125 N |

c. resting on a horizontal surface with a force of 25 N pulling up on the block.

|Fn = Fg – Fp = 100 N – 25 N = 75 N |

d. resting on a 30o incline.

|Fn = Fgcosθ = (100 N)cos30 = 87 N |

26. A 10-kg block rests on a surface. The coefficients of friction are μs = 0.3, μk = 0.15.

a. What is the maximum F|| that can act on the stationary block?

|F = Ff = μsFn = (0.3)(10 kg)(10 m/s2) = 30 N |

b. What is the force of friction when F|| = 10 N?

|Ff = 10 N |

c. What is the force of friction when F|| = 40 N?

|Ff = μkFn = (0.15)(100 N) = 15 N |

C. Force Problems—Dynamics

Kinetic Friction Lab

Time how long it takes the wood block to rise to the top of the incline (if it takes more than 1.5 s, change the angle of the incline)

a. Collect the following data.

|Hanging weight M1 |Block mass M2 (kg) |distance d |Angle θ |

|(kg) | |(m) | |

| | | | |

|Time |Trial 1 |Trial 2 |Trial 3 |tav |

|(s) | | | | |

| | | | | |

b. Calculate the following from the data.

| |Formula |Calculation |

|a |d = ½at2 | |

|mtot |m = M1 + M2 | |

|Fg-M1 |Fg-M1 = M1g | |

|Fg-M2|| |Fg-M2|| = M2gsinθ | |

|Ff |Fg-M1 – Fp – Fg-M2|| = mtota | |

|Fn |Fn = FgM2cosθ | |

|μ |Ff = μFn | |

Questions 27-37 Briefly explain your answer.

27. Two blocks of masses 2m and m are in contact on a horizontal frictionless surface. If a force F is applied to mass 2m, what is the force on mass m?

F

(A) 2 F (B) F (C) 1/2F (D) 1/3F

|D—Force is proportional to mass when acceleration is the same, since m = |

|1/3 total mass then force = 1/3F |

28. Three blocks of mass 3m, 2m and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tensions in each string?

T3 T2 T1

(A) T1 > T2 > T3 (B) T3 > T2 > T1 (C) T1 = T2 = T3

|A—T1 pulls all the mass, T2 pulls 5/6 of the mass and T3 pulls ½ of the |

|mass, ∴ T1 > T2 > T3 |

29. You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope?

(A) 0 N (B) 50 N (C) 100 N (D) 200 N

|C—Newton's third law of equal and opposite forces |

30. Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope?

(A) 0 N (B) 50 N (C) 100 N (D) 200 N

|C—Whether tied to a tree or pulled with equal force, the tension is the |

|same, 100 N |

31. You tie a rope to a tree and you and your friend each pull on the free end of the rope with forces of 100 N. What is the tension in the rope?

(A) 0 N (B) 50 N (C) 100 N (D) 200 N

|D—The combined forces on one end will generate a reaction force equal to |

|200 N |

32. In which case does the acceleration the greatest; in case 1, where a 10-kg hanging mass generates a downward force of 100 N or case 2 where a hand generates a downward force of 100 N?

(A) case 1

(B) case 2

(C) both cases are the same 100 N

|B—F = ma, where m = 50 kg + 10 kg for case 1 and 50 kg for case 2 ∴ case |

|2 where the mass is less |

33. A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why?

(A) component of the gravity parallel to the plane increases

(B) normal force exerted by the board decreases

(C) both (A) and (B)

|C—ΣF|| = Fgsinθ – μFgcosθ: as θ increases the first term increases and |

|the second term decreases |

34. A block of mass m slides down a rough incline at constant speed. If a similar block of mass ½m were placed on the same incline, its speed would

(A) be zero (B) increase (C) be constant

|C—ΣF|| = mgsinθ – μmgcosθ = ma: m cancels out of the equation ∴ m doesn't|

|affect the situation |

Questions 35-37 A block of mass m rests on the floor of an elevator. Use the following options.

(A) Fn > Fg (B) Fn = Fg (C) Fn < Fg (D) Fn = 0

35. Which is true if elevator is at constant velocity downward?

|B—Since acceleration is zero, Fnet = Fn – Fg = 0 |

|∴ Fn = Fg |

36. Which is true if the elevator is accelerating upward?

|A—Since acceleration is upward, Fnet = Fn – Fg > 0 |

|∴ Fn > Fg |

37. Which is true if the elevator is allowed to fall freely?

|D—Since acceleration is equal to -g, Fnet = Fn – Fg = -mg ∴ Fn = Fg – mg |

|= 0 |

38. A stationary 5-kg box is pulled upward by a 70-N force, Fp.

a. Draw all the forces acting on the box.

Fp = 70 N

Determine the Fg = 50 N

b. overall force acting on the box.

|Fnet = Fp – Fg |

|Fnet = 70 N – (5 kg)(10 m/s2) = 20 N |

c. acceleration of the box.

|Fnet = ma |

|20 N = (5 kg)a ∴ a = 4 m/s2 |

d. vertical velocity of the box after 2 s of acceleration.

|vt = vo + at |

|v = 0 + (4 m/s2)(2 s) = 8 m/s upward |

39. A 0.5-kg hockey puck traveling at 10 m/s on a horizontal surface slows to 2 m/s over a distance of 80 m.

a. Draw all the forces acting on the hockey puck

Fn

Ff

Determine the Fg

b. acceleration of the puck (use kinematics).

|vt2 = vo2 + 2ad |

|(2 m/s)2 = (10 m/s)2 + 2a(80 m) ∴ a = -0.6 m/s2 |

c. force of friction acting on the puck.

|Ff = ma |

|Ff = (0.5 kg)(-0.6 m/s2) = -0.3 N |

d. normal force acting on the puck.

|Fn = Fg = mg |

|Fn = (0.5 kg)(10 m/s2) = 5 N |

e. coefficient of kinetic friction.

|Ff = μFn |

|0.3 N = μ(5 N) ∴ μ = 0.06 |

40. An 80-kg skier descends a 30o slope.

a. Draw all forces acting on the skier.

Fn

Ff

Fg

Determine the

b. parallel component of the force of gravity Fg-||.

|Fg-|| = Fgsinθ |

|Fnet = (80 kg)(10 m/s2)sin30 = 400 N |

c. perpendicular component Fg-⊥.

|Fg-⊥ = Fgcosθ |

|Fg-⊥ = (80 kg)(10 m/s2)cos30 = 690 N |

d. normal force Fn.

|Fn = Fg-⊥ = 690 N |

e. force of friction Ff. (μk = 0.10)

|Ff = μFn |

|Ff = 0.10(690 N) = 69 N |

f. overall force acting on the skier parallel to the surface.

|Fnet = Fg-|| – Ff |

|Fnet = 400 N – 69 N = 330 N |

g. skier’s acceleration.

|Fnet = ma |

|330 N = (80 kg)a ∴ a = 4.1 m/s2 |

41. An elevator (m1 = 1150 kg) is suspended over a pulley which is connected to a counterweight (m2 = 1000 kg).

a. Draw all forces acting on the elevator and counterweight.

Ft-1

Ft-2

Fg-1 = 11500 N

Determine the Fg-2 = 100,000 N

b. overall force acting on the system.

|Fnet = Fg-1 – Ft-1 + Ft-2 – Fg-2 = 11500 N – 10000 N = 1500 N |

c. acceleration.

|Fnet = ma |

|1500 N = (m1 + m2)a = (2150 kg)a ∴ a = 0.70 m/s2 |

d. tension in the cable above the elevator.

|Fnet = Fg-1 – Ft-1 = m1a |

|11500 N – Ft-1 = (1150 kg)(0.70 m/s2) ∴ Ft-1 = 10700 N |

e. tension in the cable above the counterweight.

|Fnet = Ft-2 – Fg-2 = m2a |

|Ft-2 – 10000 N = (1000 kg)(0.70 m/s2) ∴ Ft-2 = 10700 N |

f. time it takes to travel 3 m starting from rest.

|d = vot + ½at2 |

|3 m = 0 + ½(0.70 m/s2)t2 ∴ t = 2.9 s |

g. speed of the elevator after traveling 3 m from rest.

|v2 = vo2 + 2ad = 0 + 2(0.70 m/s2)(3 m) = 4.2 m2/s2 |

|v = 2.0 m/s |

h. the apparent weight of a 50-kg elevator passenger.

|Fnet = Fg – Fn = ma |

|500 N – Fn = (50 kg)(0.70) ∴ Fn = 465 N |

42. A 25-kg block is pulled vertically upward by a force of 325 N.

a. Draw all the forces acting on the block to the right.

Determine the Fp = 325 N

b. overall force acting on the block.

|Fnet = Fp – Fg = 325 N – 250 = 75 N |

c. block's acceleration.

|Fnet = ma |

|75 N = (25 kg)a ∴ a = 3 m/s2 |

d. block's vertical velocity after 2 s of motion. Fg = 250 N

|vt = vo + at = 0 + (3 m/s2)(2 s) = 6 m/s |

43. Object A (mA = 10 kg) at rest on a table (μ = 0.30) is attached to B (mB = 5 kg).

a. Draw all forces acting on A and B.

Fn

Ff Ft-A

Ft-B

Fg-A

Determine the Fg-B

b. force of friction on A.

|Ff = μFn = μmAg |

|Ff = (0.30)(10 kg)(10 m/s2) = 30 N |

c. overall || force on the system.

|Fnet = Fg-B – Ff |

|Fnet = (5 kg)(10 m/s2) – 30 N = 20 N |

d. acceleration of the system.

|Fnet = (mA + mB)a |

|20 N = (10 kg + 5 kg)a ∴ a = 1.3 m/s2 |

e. tension in the rope.

|Fnet = Fg-B – Ft = mBa |

|(5 kg)(10 m/s2) – Ft = (5 kg)(1.3 m/s2) ∴ Ft = 44 N |

f. velocity of B after it has descended 0.5 m.

|vt2 = vo2 + 2ad |

|vt2 = 0 + 2(1.3 m/s2)(0.5 m) ∴ vt = 1.1 m/s |

44. A 0.50-kg hockey puck traveling at 15 m/s on a horizontal surface slows to a stop in 90 m.

a. Draw all the forces acting on the hockey puck

Fn

Ff

Determine the Fg

b. puck's acceleration (use kinematics).

|vt2 = vo2 + 2ad |

|(0 m/s)2 = (15 m/s)2 + 2a(90 m) ∴ a = -1.25 m/s2 |

c. force of friction.

|Ff = ma |

|Ff = (0.50 kg)(-1.25 m/s2) = -0.625 N |

d. normal force.

|Fn = Fg = mg |

|Fn = (0.50 kg)(10 m/s2) = 5 N |

e. coefficient of kinetic friction.

|Ff = μFn |

|0.625 N = μ(5 N) ∴ μ = 0.125 |

45. A 10-kg block slides down a 37o incline.

a. Draw all forces acting on the block. Ff

Fn

θ Fg−⊥

Fg

Fg-||

θ

Determine the

b. parallel component of the force of gravity Fg-||.

|Fg-|| = Fgsinθ |

|Fg-|| = (10 kg)(10 m/s2)sin37 = 60 N |

c. perpendicular component Fg-⊥.

|Fg-⊥ = Fgcosθ |

|Fg-⊥ = (10 kg)(10 m/s2)cos37 = 80 N |

d. normal force Fn.

|Fn = Fg-⊥ = 80 N |

e. force of friction Ff. (μk = 0.25)

|Ff = μFn |

|Ff = 0.25(80 N) = 20 N |

f. overall force acting on the block.

|Fnet = Fg-|| – Ff |

|Fnet = 60 N – 20 N = 40 N |

g. block's acceleration.

|Fnet = ma |

|40 N = (10 kg)a ∴ a = 4 m/s2 |

46. A 1.5-kg weight is suspended over a pulley which is connected to a 1-kg weight

a. Draw all forces acting on the 1.5-kg and 1-kg weights.

Ft-1 Ft-1.5

Fg-1 Fg-1.5

Determine the

b. overall force acting on the system.

|Fnet = Fg-1.5 – Ft-1.5 + Ft-1 – Fg-1 |

|Fnet = (m1.5 – m1)g = (1.5 kg – 1 kg)(10 m/s2) = 5 N |

c. acceleration.

|Fnet = ma |

|5 N = (2.5 kg)a ∴ a = 2 m/s2 |

d. tension in the cable above the 1.5-kg weight.

|Fnet = Fg-1.5 – Ft-1.5 = m1.5a |

|15 N – FT-1.5 = (1.5 kg)(2 m/s2) ∴ Ft-1.5 = 12 N |

e. tension in the cable above the 1-kg weight.

|Fnet = Ft-1 – Fg-1 = m1a |

|Ft-1 – 10 N = (1 kg)(2 m/s2) ∴ Ft-1 = 12 N |

f. time it takes to travel 2.25 m starting from rest.

|d = vot + ½at2 |

|2.25 m = 0 + ½(2 m/s2)t2 ∴ t = 1.5 s |

D. Force Problems—Statics

Force Table Lab

Determine the missing force and angle using spring scales and compare the results to the calculated values.

a. Collect the following data.

| | | | |

|Experiment |Scale A |Scale B |Scale C |

| |( |FA |( |

|Formula |Calculation |

|FAx |Ax = Acosθ | | | |

|FBx |Bx = Bcosθ | | | |

|FCx |Ax + Bx + Cx = 0 | | | |

|FAy |Ay = Asinθ | | | |

|FBy |By = Bsinθ | | | |

|FCy |Ay + By + Cy = 0 | | | |

|FC |C = (Cx2 + Cy2)½ | | | |

|θ |tanθ = Cy/Cx | | | |

c. Calculate the percent difference between the measured values of (part a) and theoretical values of (part b).

|Experiment |1 |2 |3 |

|Formula |Calculation |

|Fc |%Δ =100|Δ|/C | | | |

|θ | | | | |

47. The 100-N block is stationary and μs = 0.40.

a. What is the minimum weight W?

|W + Ff = 75 N |

|W = 75 – μFg = 75 – (0.40)(100) = 35 N |

b. What is the maximum weight W?

|W = 75 + Ff |

|W = 75 + 40 = 115 N |

48. A 50-kg box is anchored to the ceiling by two cords.

45o 30o

a. Use ΣFx = 0 to write an expression for the tension in the right cable in terms of the tension in the left cable.

|TLcos135 + TRcos30 = 0 |

|TL(-0.707) + TR(0.866) = 0 ∴ TR = TL(0.816) |

b. Use ΣFy = 0 to solve for the tension in the left cable.

|TLsin135 + TRsin30 + mgsin-90 = 0 |

|TL(0.707) + TL(0.816)(0.500) + (30 kg)(10 m/s2)(-1) = 0 |

|TL(1.12) = 300 N ∴ TL = 269 N |

c. Substitute the left tension into the expression from (a).

|TR = TL(0.816) = (269 N)(0.816) = 220 N |

49. A 50-kg box is anchored to the ceiling and wall by cords.

60o

TL

TR

a. Label each force in the vector sum diagram.

Fg TL

60o

TR

b. Calculate the tension in the ceiling cord (TL).

|sin60 = Fg/TL = 500 N/TL ∴ TL = 577 N |

c. Calculate the tension in the wall cord (TR).

|tan60 = Fg/TR = 500 N/TR ∴ TR = 289 N |

50. A 50-kg box is anchored to the ceiling by two cords, which form a 90o angle between them.

53o 37o

TL TR

b. Calculate TL.

|sin53 = TL/Fg = TL/500 N ∴ TL = 400 N |

c. Calculate TR.

|cos53 = TR/Fg = TR/500 N ∴ TR = 300 N |

Practice Multiple Choice (No calculator)

Briefly explain why the answer is correct in the space provided.

1. A 6-N force and an 8-N force act on a point. The resulting force is 10 N. What is the angle between the two forces?

(A) 0o (B) 30o (C) 45o (D) 90o

|D—The vector sum forms a 6-8-10 (3-4-5) triangle, which is a right |

|triangle, where the 3-4 sides make 90o |

2. A 60-kg physics student would weigh 1560 N on the surface of planet X. What is the acceleration due to gravity on the surface of planet X?

(A) 2 m/s2 (B) 10 m/s2 (C) 15 m/s2 (D) 26 m/s2

|D—Fg = mg |

|1560 N = (60 kg)g ∴ g = 26 m/s2 |

3. A 400-N girl standing on a dock exerts a force of 100 N on a 10,000-N sailboat as she pushes it away from the dock. How much force does the sailboat exert on the girl?

(A) 25 N (B) 400 N (C) 100 N (D) 10000 N

|C—Newton's third law states that for every force there is an equal and |

|opposite reaction force |

4. What force is needed to keep a 60-N block moving across asphalt (μ = 0.67) in a straight line at a constant speed?

(A) 40 N (B) 51 N (C) 60 N (D) 120 N

|A—W/o acceleration then F = Ff. |

|Ff = μFn = μmg → Ff = (2/3)(60 N) = 40 N |

5. If F1 is the force exerted by the earth on the moon and F2 is the force exerted by the moon on the earth, then which of the following is true?

(A) F1 equals F2 and in the same direction

(B) F1 equals F2 and in the opposite direction

(C) F1 is greater than F2 and in the opposite direction

(D) F2 is greater than F1 and in the opposite direction

|B—Newton's third law states that for every force there is an equal, but |

|opposite reaction force |

6. How far will a spring (k = 100 N/m) stretch when a 10-kg mass hangs vertically from it?

(A) 1 m (B) 1.5 m (C) 5 m (D) 10 m

|A—Fs = kx → mg = kx |

|(10 kg)(10 m/s2) = (100 N/m)x ∴ x = 1 m |

7. Compared to the force needed to start sliding a crate across a rough level floor, the force needed to keep it sliding once it is moving is

(A) less (B) greater (C) the same

|A—The coefficient μk is less when the object is sliding (μk < μs) ∴ it |

|takes less force to keep it sliding |

8. Which vector diagram best represents a cart slowing down as it travels to the right on a horizontal surface?

(A) (B) (C) (D)

[pic]

|B—Slowing down ∴ the force of friction, Ff, must be greater than the |

|push/pull force, F |

9. A skier on waxed skis (μk = 0.05) is pulled at constant speed across level snow by a horizontal force of 40 N. What is the normal force exerted on the skier?

(A) 700 N (B) 750 N (C) 800 N (D) 850 N

|No acceleration then F = Ff = μFn |

|40 N = (0.05)Fn ∴ Fn = 800 N |

10. Two 0.60-kg objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. A 0.30-kg mass is added on top of one of the 0.60-kg objects and the objects are released.

[pic]

The acceleration of the system of objects is most nearly

(A) 10 m/s2 (B) 6 m/s2 (C) 3 m/s2 (D) 2 m/s2

|C—Fnet = ma (Fnet = extra weight and m = total mass) |

|(0.30 kg)(10 m/s2) = (.30 + .60 + .60 kg)a ∴ a = 2 m/s2 |

11. The diagram shows a 4.0-kg object accelerating at 10 m/s2 on a rough horizontal surface.

[pic]

What is the frictional force, Ff, acting on the object?

(A) 5 N (B) 10 N (C) 20 N (D) 40 N

|B—Fnet = F – Ff = ma |

|50 N – Ff = (4.0 kg)(10 m/s2) ∴ Ff = 10 N |

12. Two blocks are pushed along a horizontal frictionless surface by a force, F = 20 N to the right.

F

The force that the 2-kg block exerts on the 3-kg block is

(A) 8 N to the left (B) 8 N to the right

(C) 10 N to the left (D) 12 N to the right

|A—Force is proportional to the mass being pushed ∴ F2/F = 2/5 ∴ F2 = |

|2/5F = 2/5(20 N) = 8 N (left) |

Questions 13-15 A box of mass m is on a ramp tilted at an angle of θ above the horizontal. The box is subject to the following forces: friction (Ff), gravity (Fg), pull (Fp) and normal (Fn). The box is moving up the ramp.

Fn Fp

Ff Fg

θ

13. Which formula is correct for Fn?

(A) Fn = Fg (B) Fn = Fgcosθ

(C) Fn = Fgsinθ (D) Fn = Fgcosθ

|D—Fn equals the component of gravity that is perpendicular to the |

|surface: Fgcosθ |

14. Which formula is correct for Ff if the box is accelerating?

(A) Ff = Fp (B) Ff = Fp – Fgsinθ

(C) Ff = Fgcosθ (D) Ff = μFn

|D—Ff = μFn |

15. Which formula is correct for Fp if the box is moving at constant velocity?

(A) Fp = Ff (B) Fp = Ff + Fgsinθ

(C) Fp = ma - Ff (D) Fp = Ff – Fgsinθ

|B—Fnet = ma = 0 |

|Fp – Ff – Fgsinθ = 0 ∴ Fp = Ff + Fgsinθ |

16. A rope supports a 3-kg block. The breaking strength of the rope is 50 N. The largest upward acceleration that can be given to the block without breaking the rope is most nearly

(A) 6 m/s2 (B) 6.7 m/s2 (C) 10 m/s2 (D) 15 m/s2

|B—Fnet = Fp – Fg = ma |

|50 N – (3 kg)(10 m/s2) = (3 kg)a ∴ a = 20/3 = 6.7 m/s2 |

17. A block of mass 3m can move without friction on a horizontal table. This block is attached to another block of mass m by a cord that passes over a frictionless pulley. The masses of the cord and the pulley are negligible,

[pic]

What is the acceleration of the descending block?

(A) Zero (B) ¼ g (C) ⅓ g (D) ⅔ g

|B—Fnet = ma (Fnet is Fg on m, and m = total mass = 4 m) |

|mg = (4m)a ∴ a = ¼ g |

18. A 5-kg block lies on a 5-m inclined plane that is 3 m high and 4 m wide. The coefficient of friction between the plane and the block is 0.3.

What minimum force F will pull the block up the incline?

(A) 12 N (B) 32 N (C) 42 N (D) 50 N

|C—Fnet = Ff + Fg-|| |

|Fnet = μ(4/5mg) + 3/5mg = (0.3)(40 N) + 30 N = 42 N |

19. A 10-kg block is pulled along a horizontal surface at constant speed by a 50-N force, which acts at a 37o angle with the horizontal.

What is the coefficient of friction μ?

(A) 1/3 (B) 2/5 (C) 1/2 (D) 4/7

|D—Fnet = Ff = μFn |

|(50 N)cos37 = μ[(10 kg)(10 m/s2) – (50 N)sin37] ∴ μ = 4/7 |

20. A 10-kg cart moves without frictional loss on a level table. A 50-N force pulls horizontally to the right (F1) and a 40-N force at an angle of 60° pulls on the cart to the left (F2).

F2

F1

What is the horizontal acceleration of the cart?

(A) 1 m/s2 (B) 2 m/s2 (C) 3 m/s2 (D) 4 m/s2

|C—Fnet = F1 – F2cos60 = ma (cos60 = ½) |

|50 N – 40 N(½) = (10 kg)a ∴ a = 3 m/s2 (to the right) |

21. A push broom of mass m is pushed across a rough floor by a force T directed at angle θ. The coefficient of friction is μ.

The frictional force on the broom has magnitude

(A) μ(mg + Tsinθ) (B) μ(mg - Tsinθ)

(C) μ(mg + Tcosθ) (D) μ(mg - Tcosθ)

|A—Ff = μFn |

|Ff = μ(mg + Tsinθ) |

Questions 22-23 A block of mass m is accelerated across a rough surface by a force F that is exerted at an angle φ with the horizontal. The frictional force is f.

[pic]

22. What is the acceleration of the block?

(A) F/m (B) Fcosφ/m

(C) (F – f)/m (D) (Fcosφ – f)/m

|D—Fnet = Fcosφ – f = ma |

|a = (Fcosφ – f)/m |

23. What is the coefficient of friction between the block and the surface?

(A) f/mg (B) mg/f

(C) (mg – Fcosφ)/f (D) f/(mg – Fsinφ)

|D—f = μFn = μ(mg – Fsinφ) |

|μ = f/(mg – Fsinφ) |

24. A horizontal force F pushes a block of mass m against a vertical wall. The coefficient of friction between the block and the wall is μ.

What minimum force, F, will keep the block from slipping?

(A) F = mg (B) F = μmg

(C) F = mg/μ (D) F = mg(1 – μ)

|C—Fnet = Ff – Fg = 0 |

|0 = μFn – mg = μF – mg ∴ F = mg/μ |

25. Two blocks of masses M and m, with M > m, are connected by a light string. The string passes over a frictionless pulley of negligible mass so that the blocks hang vertically. The blocks are then released from rest. What is the acceleration of the block of mass M?

(A) g (B) (M – m)g/M

(C) (M + m)/Mg (D) (M – m)g/(M + m)

|D—Fnet = ma |

|(M – m)g = (M + m)a ∴ a = (M – m)g/(M + m) |

26. The momentum vector for an object is represented below.

The object strikes a second object that is initially at rest.

Which set of vectors may represent the momentum of the two objects after the collision?

(A) (B)

(C) (D)

|C—When you place the two vectors tail to tip, only c will equal the |

|original vector. |

Questions 27-28 A 100-N weight is suspended by two cords.

[pic]

27. The tension in the ceiling cord is

(A) 50 N (B) 100 N (C) 170 N (D) 200 N

|D—TL, Fg and TR make a 30-60-90 triangle, where |

|sin30 = Fg/TR ∴ TR = (100 N)/sin30 = 100 N/½ = 200 N |

28. The tension in the wall cord is

(A) 50 N (B) 100 N (C) 170 N (D) 200 N

|C—tan30 = Fg/TL |

|∴ TL = (100 N)/tan30 = 100 N/(1/√3) = 170 N |

Practice Free Response

1. Block A (8 kg) is pulled along a frictionless table by block B (2 kg) hanging over the table's edge.

a. Label all the forces acting on the two blocks.

| Fn |

| |

|Ff Ft-A |

| |

|Ft-B |

|Fg-A |

| |

| |

| |

|Fg-B |

b. Determine the acceleration of the system.

|Fnet = ma |

|20 N = (8 kg + 2 kg)a ∴ a = 2 m/s2 |

c. Determine the tension on the 8-kg block.

|Ft = ma = (8 kg)(2 m/s2) = 16 N |

d. Determine the tension on the 2-kg mass.

|Fg – Ft = ma |

|(2 kg)(10 m/s2) – Ft = (2 kg)(2 m/s2) ∴ Ft = 16 N |

2. An empty sled of mass 25 kg slides down a snowy hill with a constant speed of 2.4 m/s. The slope of the hill is inclined at an angle of 15° with the horizontal.

a. Calculate the time it takes the sled to go 21 m down the slope.

|d = vt |

|21 m = (2.4 m/s)t ∴ t = 8.75 s |

b. Label all the forces acting on the sled.

| Fn |

|Ff |

| |

| |

| |

|Fg |

c. Calculate the force of friction.

|Fnet = Fgsinθ – Ff = 0 |

|Ff = (25 kg)(10 m/s2)sin15o = 65 N |

d. Calculate the coefficient of friction.

|Ff = μFn = μ(Fgcosθ) |

|65 N = μ(25 kg)(10 m/s2)cos15o ∴ μ = 0.27 |

e. The sled reaches the bottom of the hill and continues on horizontal ground. Graph velocity vs. time for the sled until it stops (Assume the same μ). Indicate the time t when the sled reaches the horizontal ground.

velocity

t time

3. A 4700 kg truck carrying a 900 kg crate is traveling at 25 m/s to the right along a straight, level highway. The driver then applies the brakes, and as it slows down, the truck travels 55 m in the next 3.0 s. The crate does not slide.

a. Calculate the magnitude of the acceleration of the truck, assuming it is constant.

|d = vot + ½at2 |

|55 m = (25 m/s)(3.0 s) + ½a(3.0 s)2 ∴ a = -4.4 m/s2 |

b. Label all the forces acting on the crate during braking.

Fn

Ff

Fg

c. Calculate the minimum coefficient of friction between the crate and truck that prevents the crate from sliding.

|F = Ff = μFn |

|ma = μmg ∴ μ = a/g = 4.4 m/s2/10 m/s2 = 0.44 |

Now assume the bed of the truck is frictionless, but there is a spring of spring constant 9200 N/m attaching the crate to the truck. The truck is initially at rest.

[pic]

d. If the truck and crate have the same acceleration, calculate the extension of the spring as the truck accelerates from rest to 25 m/s in 10 s.

|a = (v – vo)/t = 25 m/s/10 s = 2.5 m/s2 |

|F = Fs = kx → ma = kx |

|x = ma/k = (900 kg)(2.5 m/s2)/(9200 N/m) = 0.24 m |

e. What happens to the spring when the truck reaches 25 m/s and stops accelerating? Explain your reasoning.

|The spring shortens. Without acceleration, the spring force is not |

|needed to keep the crate on the truck bed so the spring will go to its |

|unstretched length. |

4. A 28.0-kg block rests on a table

(μs = 0.450, μk = 0.320). The block

is connected to an empty 1.35-kg

bucket by a cord running over a

frictionless pulley. Sand is gradually

added to the bucket until the system

just begins to move.

a. Calculate the mass of sand added to the bucket.

|Fg = Ff → (mbucket + msand)g = μsmblockg |

|1.35 kg + msand = (0.450)(28.0 kg) ∴ msand = 11.25 kg |

b. Calculate the acceleration of the system.

|Fnet = Fgbucket + Fgsand – Ffblock = mtotala |

|(mbucket + msand – μkmblock)g = (mbucket + msand + mblock)a |

|a = [12.6 kg – (0.320)(28.0 kg)]10 m/s2]/40.6 kg = 0.90 m/s2 |

5. A 7650-kg helicopter accelerates upward at 0.80 m/s2 while lifting a 1250-kg frame at a construction site.

a. What is the lift force of the helicopter rotors?

|Fp = mg + ma = m(g + a) |

|Fp = (7650 kg + 1250 kg)(10.80 m/s2) = 96,100 N |

b. What is the tension in the cable (ignore its mass) that connects the frame to the helicopter?

|Fnet = ma → Ft – mg = ma |

|Ft = m(a + g) = (1250 kg)(0.80 m/s2 + 10 m/s2) = 13,500 N |

c. What force does the cable exert on the helicopter?

|Fp – Fg –Ft = ma |

|96,100 N – 76500 N – Ft = (7650 kg)(0.80 m/s2) |

|Ft = 13,500 N |

6. A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling. One end of the rope is held by Student A of mass 70 kg, who is at rest on the floor. The opposite end of the rope is held by Student B of mass 60 kg, who is suspended at rest above the floor.

a. Draw and label all the forces on Student A and on Student B.

TB

TA

Fg-B

Fg-A

b. Calculate the magnitude of the force exerted by the floor on Student A.

|Fn = Fg-A – Fg-B = mAg – mBg = (mA – mB)g |

|Fn = (70 – 60)(10) = 100 N |

Student B now climbs up the rope at a constant acceleration of 0.25 m/s2 with respect to the floor.

c. Calculate the tension in the rope while Student B is accelerating.

|Fnet = TB – Fg-B = ma |

|TB = mBg + mBa = mB(g + a) = (60)(10 + 0.25) = 615 N |

d. As Student B is accelerating, is Student A pulled upward off the floor? Justify your answer.

|No, the upward force TA = 615 N < the downward force Fg-A = 700 N ∴ the |

|net force on student A is downward. |

e. With what minimum acceleration must Student B climb up the rope to lift Student A upward off the floor?

|Fnet = TB – FgB = mBa (TB = TA = mAg) |

|mAg – mBg = mBa |

|a = (mA – mB)g/mB = (70 – 60)10/60 = 1.7 m/s2 |

f. How could student B's acceleration up the rope (measured from the floor) be increased?

|Anchor student A with additional weight. |

7. A 100-kg box is anchored to the ceiling by two cords.

50o 25o

Determine the tension in each cord.

|TLcos130 + TRcos25 = 0 |

|TL(-0.643) + TR(0.906) = 0 ∴ TR = TL(0.710) |

|TLsin130 + TRsin25 + mgsin-90 = 0 |

|TL(0.766) + TL(0.710)(0.423) + (100 kg)(10 m/s2)(-1) = 0 |

|TL(1.07) = 1000 N ∴ TL = 938 N |

|TR = TL(0.710) = (938 N)(0.710) = 666 N |

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50 kg

m

2m

3m

m

2m

m2

m1

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