California State Polytechnic University, Pomona



PHY 131 Ch 1 to 5 Name:

1.

|A stone is dropped from a cliff. The |[pic] |

|graph that best represents its motion | |

|while it falls is: | |

|III | |

One object is thrown vertically upward with an initial velocity of 100 m/s and another object with an initial velocity of 10 m/s. The maximum height reached by the first object will be _______ that of the other.

| a = ∆v / ∆t | a = ∆v / ∆t |

|-10 = 0 – 100 / t |-10 = 0 – 10 / t |

|t = 10 sec |t = 1 sec |

| | |

|y = ½ a t2 + vo t |y = ½ a t2 + vo t |

|y = ½-10 t2 + 100t |y = ½-10 t2 + 10t |

|y = 500 m |y = 5 m |

An object placed on an equal-arm balance requires 12 kg to balance it. When placed on a spring scale, the scale reads 12 kg. Everything is now transported to the Moon where free-fall acceleration is 1/6 that on Earth. The new readings of the balance and spring scale (respectively) are:

-kx = mg (gmoon = 1.66 m/s2) Ans: 12 kg, 2 kg

Two blocks weighing 250 N and 350 N

respectively, are connected by a string

that passes over a massless pulley as shown.

The tension in the string is:

FNet = mT a

350 – 250 = (25+35) a

a = 1.7 m/s2

Accel down Accel UP

FT = 35kg (10-1.7) OR 25kg (10+1.7)

FT = 290 N

An object is thrown straight down with an initial speed of 4 m/s from a window which is 8 m above the ground. The time it takes the object to reach the ground is:

y = ½ a t2 + vot + yo

0 = ½-10 t2 - 4t + 8

t = 0.93 sec

A Ferris wheel with a radius of 8.0 m makes 1 revolution every 10 s. As he passed the top of the path he releases a ball. How far from the point on the ground directly under the release point does the ball land?

|v = 2πr / T |h = ½ a t2 |vx = ∆x / ∆t |

|v = 2π8 / 10 |2r = ½10 t2 |5 = ∆x / 1.8 |

|v = 5 m/s |tbottom = 1.8 sec |∆x = 9 m |

A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45°. With what speed was it thrown?

|y = ½ a t2 | a = ∆vy / ∆t |

|20 = ½-10 t2 |-10 = (sin45v – 0)/2 |

|t = 2 sec |v = 28 m/s |

A large cannon is fired from ground level over level ground at an angle of 30° above the horizontal. The muzzle speed is 980 m/s. The projectile will travel what horizontal dist before striking the ground?

|y = ½ a t2 + vo t + yo | vx = ∆x /∆t |

|0 = ½-10t2 + sin30(980) t + 0 |cos30 (980) = ∆x /98 |

|t = 98 sec |∆x = 83000 m |

A ferry boat is sailing at 12 km/h 30° W of N with respect to a river that is flowing at 6.0 km/h E. As observed from the shore, the ferry boat is sailing:

| | |vx |vy |

|No x component | | | |

|Due North | | | |

| |River |6 |0 |

| |Ferry |-sin30 12 |cos30 12 |

| |Sum |0 |10.4 |

When a 40-N force, parallel and directed up a ramp inclined by 30° above the horizontal, the acceleration of the crate is 2.0 m/s2, up the incline. The mass of the crate is:

| FNet = mT a |[pic] |

|40 – sin30 mg = m (2) | |

|m = 5.7 kg | |

A 25-kg crate is pushed across a frictionless horizontal floor with a force of 20 N, directed 20° below the horizontal. The accel of the crate is:

FNet = mT a

cos20 (20)= 25 a

a = 0.75 m/s2

|A 5-kg block is suspended by a rope from the ceiling of |FT = m (g + a) |

|an elevator that accelerates downward at 3.0 m/s2. The |FT = 5 (10 + 3) |

|tension force of the rope on the block is: |FT = 65 N, up |

The position of an object as a function of time is given in meters by x = (at +bt2) i + (ct) j. What is its velocity as a function of time? v = dx / dt

v = (a + 2bt) i + (c) j

An object is thrown vertically into the air. Which of the following five graphs represents the velocity (v) of the object as a function of the time (t)? (Up is positive)

[pic]III

A projectile is shot vertically upward with a given initial velocity. It reaches a maximum height of 100 m. If on a second shot the initial velocity is doubled, then the projectile will reach a maximum height of:

|y = ½ a t2 | a = ∆v / ∆t |

|100 = ½-10 t2 |-10 = 0 – 90 / t |

|t = √20 |t = 9 sec |

| | |

|a = ∆v / ∆t |y = ½ a t2 + vo t |

|-10 = 0 – vo / √20 |y = ½-10 t2 + 90t |

|vo = 45 m/s |y = 400 m |

|A boat is traveling upstream at 14 mph with respect |v2 = vx2 + vy2 |

|to a river that is flowing at 6 mph. A man runs |v2 = (14-6)2 + 62 |

|directly across the boat, from one side to the other,|v = 10 mph |

|at 6 mph. The speed of the man with respect to the | |

|ground is: | |

A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along the circumference of the circle, in 5.0 s. The magnitude of her acceleration is:

|¼(2πr) = 25m |a = v2 / r |

|r = 50/π |a = (25/5)2 / 15.9 |

|r = 15.9 m |a = 1.6 m/s2 |

|A dart is thrown horizontally at 20 m/s. It hits 0.1 s |y = ½ a t2 |

|later. What is the distance the dart drops during its |y = ½10 .12 |

|path? |y = 0.05 m |

Two forces, one with a magnitude of 3 N and the other with a magnitude of 5 N, are applied to an object. For which orientation of the shown forces shown is the magnitude of the acceleration of the object the least?

|[pic] |We definitely know: |

| | |

| |IV is Greatest, and|

| | |

| |I is the Least |

|A baseball is hit straight up and is caught by the catcher |y = ½ a ttop2 |

|2.0 s later, at the same height at which it left the bat. |y = ½10 12 |

|The maximum height of the ball during this interval is: |y = 5 m |

|A crate rests on a horizontal surface and a woman pulls on it with a 10-N |

|force. Rank the situations shown below according to the magnitude of the |

|normal force exerted by the surface on the crate, least to greatest. |

| |[pic] |

|3, 2, 1 | |

|A car drives in a straight line on a level road with|tan θ = 3/10 |

|a constant acceleration of 3 m/s2. A ball is |θ = 17° |

|suspended by a string from the ceiling of the car; |[pic] |

|the ball does not swing, being at rest with respect | |

|to the car. What angle does the string make with the| |

|vertical? | |

When a 40-N force, parallel and directed up a ramp inclined by 30° above the horizontal, the acceleration of the crate is 2.0 m/s2, down the incline. The mass of the crate is:

| FNet = mT a |[pic] |

|40 – sin30 mg = m (-2) | |

|m = 13.8 kg | |

|A 90-kg man stands in an elevator that is moving up |FW = m (g + a) |

|at a constant speed of 5.0 m/s. The force exerted by |FW = 90 (10 + 0) |

|him on the floor is about: |FW = 900N |

|A massless rope passes over a massless pulley suspended |

|from the ceiling. A 4-kg block is attached to one end and |[pic] |

|a 5-kg block is attached to the other end. The | |

|acceleration of the 5-kg block is: | |

|FNet = mT a | |

|5g – 4g = (4+5) a | |

|a = g/9 m/s2 | |

| |[pic] |

| | |

|An object is dropped from rest. Which of the | |

|five following graphs correctly represents its| |

|motion? The positive direction is taken to be | |

|upward. Take careful note of the axes. | |

An object is thrown vertically upward with a certain initial velocity in a world where the acceleration due to gravity is 19.6 m/s2. The height to which it rises is ____ that to which the object would rise if thrown upward with the same initial velocity on the Earth. half

| a = ∆v / ∆t | a = ∆v / ∆t |

|-10 = 0 – 100 / t |-20 = 0 – 100 / t |

|t = 10 sec |t = 5 sec |

|y = ½ a t2 + vo t |y = ½ a t2 + vo t |

|y = ½-10 t2 + 100t |y = ½-20 t2 + 100t |

|y = 500 m |y = 250 m |

A short 10-g string is used to pull a 50-g toy across a frictionless horizontal surface. If a 0.030 N force is applied horizontally to the free end of the string, the force of the string on the toy, at the other end, is:

Newton’s 3rd law: 0.030 N

A motor boat can travel at 10 km/h in still water. A river flows at 5 km/h west. A boater wishes to cross from the south bank to a point directly opposite on the north bank. At what angle must the boat be headed?

|tan θ = 5 / √75 | |vx |vy |

|θ = 30° | | | |

| |River |5 |0 |

| |Boat |-5 |102 = -52 + vy2 |

| |Sum |0 |vy = √75 |

|A stone is tied to a 0.50-m string and whirled at a constant|[pic] |

|speed of 4.0 m/s in a vertical circle. The acceleration at | |

|the bottom of the circle is: a = v2/r = 32m/s2, up | |

|FT – mg = mv2/r | |

|FT = 5mg | |

A boy on the edge of a vertical cliff 20 m high throws a stone horizontally outwards with a speed of 20 m/s. It strikes the ground at what horizontal distance from the foot of the cliff?

|y = ½ a t2 |vx = ∆x / ∆t |

|20 = ½10 t2 |20 = xo – 0 / 2 |

|t = 2 sec |xo = 40 m |

A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be: F = m a a = ∆v /∆t

8 = 16 a ½ = vf – 0 / 4

a = ½ m/s2 vf = 2 m/s

A 32-N force, parallel to the incline, is required to push a certain crate at constant velocity up a frictionless incline that is 30° above the horizontal. The mass of the crate is:

| FNet = mT a |[pic] |

|32 – sin30 mg = m (0) | |

|m = 6.4 kg | |

A stone is thrown vertically upward with an initial speed of 19.5 m/s. It will rise to a maximum height of:

| a = ∆v / ∆t |y = ½ a t2 + vo t |

|-10 = 0 – 19.5 / t |y = ½-10 t2 + 19.5t |

|t = 1.95 sec |y = 19 m |

|A 100 gram pendulum bob is held at an|[pic] |tan θ = 2/1 |

|angle θ from the vertical by a 2 N | |θ = 63° |

|horizontal force F. The tension in | | |

|the string is: | |FT2 = 22 + 12 |

| | |FT = √5 |

|Two forces are applied to a 5.0-kg crate; one is 6.0 N|a2 = 6/52 + 8/52 |

|to the north and the other is 8.0 N to the west. The |a = 2 m/s2 |

|magnitude of the acceleration of the crate is: | |

|A particle goes from |(3 - -2), (-1 – 3), (4 - 1) |

|x = –2 m, y = 3 m, z = 1 m to |(5, -4, 3) |

|x = 3 m, y = –1 m, z = 4 m. | |

|Its displacement is: | |

|(45%) Below is a ramp inclined by 30.0 degrees, initially at rest. A Force, F = 30.0 N, |[pic] |

|is applied by a rocket (unlimited fuel). The 2.00 block will travel another 3 meters up | |

|the ramp before going over the edge. The coordinates of the edge are (0, 0), up ( + and | |

|right ( +. | |

|(a) What is the normal 2% 2% 8% | |

|force exerted by the FN = cos30(20) + F sin30 | |

|ramp on the block? FN = 32 Newtons | |

|(b) If a landing pad is 1 meter below | |

|the edge, how far will the block travel in the horizontal direction? | |

| |On Ramp |Free Fall with Rocket |

|FNet = mT a | | |

|8% 3% 2% 1% | | |

|Fcos30 – sin30 mg = m a | | |

|26 - 10 = 2 (a) | | |

|a = 8 m/s2 | | |

| | | |

|x = ½ a t2 2% | | |

|3 = ½ 8 t2 | | |

|t = 0.866 sec | | |

| |a = ∆v / ∆t |Fx = m ax |

| |8 = vf – 0 / .866 |30 = 2 (ax) |

| |vf = 6.93 m/s 2% |ax = 15 m/s2 5% |

| |vx = cos30(6.93) |vy = sin30(6.93) |y = ½ a t2 + vo t + yo |x = ½ a t2 + vo t + yo |

| |vx = 6 m/s 1% |vy = 3.46 m/s 1% |1% 1% 1% 1% |1% 1% 1% 1% |

| | | |0 = ½-10t2 + 3.46 t + 1 |x = ½ 15 .912 + 6(.91) + 0 |

| | | |t = 0.91 sec |x = 13.1 meters |

| | | |[pic] |

| | | | |

| | | | |

|F = 20 N | | | |

| |On Ramp |Free Fall | |

|(a) 2% 2% 8% | | | |

|FN = 20N + sin30 F | | | |

|FN = 30 Newtons | | | |

| | | | |

|FNet = mT a | | | |

|6% 1% | | | |

|Fcos30 = m a | | | |

|17.3 = 2 (a) | | | |

|a = 8.66 m/s2 | | | |

| |x = ½ a t2 2% | Fx = m ax | |

| |3 = ½ 8.66 t2 |cos30 F = 2 (ax) | |

| |t = 0.832 sec |ax = 8.66 m/s2 5% | |

| |a = ∆v / ∆t | Fy = m ax |y = ½ at2 + vot + yo |x = ½ a t2 + vo t + yo |

| |8.66 = vf – 0 / .832 |7% 1% 1% |1% 1% 1% |2% 1% 1% 1% |

| |vf = 7.21 m/s 2% |sin30F + mg = 2 (ay) |0 = ½-15t2 + 0 + 1 |x = ½ 8.66 .3652 + 7.21(.365) + 0 |

| | |ay = -15 m/s2 |t = 0.365 sec |x = 3.21 meters |

F = 8.00 N

|(a) 2% 2% 8% |End of Ramp |Free Fall |[pic] |

|FN = 20cos30 + sin30 F | | | |

|FN = 21.6 Newtons | | | |

| | | | |

|On Ramp | | | |

|(b) FNet = mT a | | | |

|2% 3% 8% 1% | | | |

|mg sin30 - Fcos30 = m a | | | |

|3.07 = 2 (a) | | | |

|a = 1.5 m/s2 | | | |

| | | | |

|x = ½ a t2 2% | | | |

|3 = ½ 1.54 t2 | | | |

|t = 2.0 sec | | | |

| |a = ∆v / ∆t |Fx = m ax | |

| |1.5 = vf – 0 / 2.0 |-8 = 2 (ax) | |

| |vf = 3.0 m/s 2% |ax = -4 m/s2 5% | |

| |vx = cos30(3) |y = ½ a t2 + vo t + yo | |

| |vx = 2.6 m/s 1% |1% 1% 1% 1% | |

| | |0 = ½10t2 + 1.5 t - 1 | |

| |vy = sin30(3) |t = 0.321 sec | |

| |vy = 1.5 m/s 1% | | |

| | |x = ½ a t2 + vo t + yo | |

| | |1% 1% 1% 1% | |

| | |x = ½ -4 .3212 + 2.6(.321) + 0 | |

| | |x = 0.64 meters | |

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