Practice Problems for TEST II, PHY2020



Practice Problems for TEST II, PHY2020

March 5 (Wednesday) in class, chapters 7-11

Problems from Chap. 7

1. If you want an elevator (total mass of elevator plus max load of occupants = 2000 kg) to go up a 50 story (200 m) tall building in 1 minute, about what power (in units of horsepower, where 1 hp = 746 watts) motor do you need?

Work = F times displacement, where in general only the component of the force in the direction of the displacement counts – i. e. this is a VECTOR product, so Work = |F| |d| cosθ, where “ |F| “ is the notation for the magnitude of the Force vector, also similarly for |d| is the magnitude of the displacement vector, and θ in the angle between vector F and vector d. All of that said, the force to lift the elevator is in the direction of the displacement, so cos(0) = 1. Work = 2000 kg * 9.8 m/s2 * 200 m, since we don’t want to accelerate the mass, just lift it at constant velocity. So power = Work done divided by the time it took to do the work, so P = W/60 s = 65,300 Watts = 88 hp. (A really big motor – probably no elevator goes that fast.)

2. An object with mass = 1 kg has how much potential energy (units of Joules) at a height of 100 m in earth’s gravitational field.

EPotential = mgh = 1kg * 9.8m/s2 * 100 m = 980 J

3. Using all of this potential energy from 2.) just above and converting into Kinetic Energy after the object falls 99.9999 m (i. e. just before impact), how fast is the object moving after it falls the ~ 100 m? (Units of m/s)

980 J = ½ mv2, so v= (2 * 980 J / 1 kg)1/2 = 44.3 m/s

-do the same calculation using 2a(y-y0) = vf2 – v02 and compare your answer (remember that v0=0)

2 * 9.8 m/s2 (100 m) = vf2 → vf = 44.3 m/s

4. The energy stored in a compressed spring that has spring constant ‘k’ and is compressed a distance ‘x’ from its relaxed state is ½ kx2. If a spring has k=1000 N/m and x is 0.1 m, how fast is a 0.01 kg object going after it has been shoved by this compressed spring? Units of m/s (convert the ½ kx2 into kinetic energy)

turn spring energy (1/2 k x2 = ½ 1000 N/m * (0.1 m)2 = 5 J) into kinetic energy =

½ mv2, so v = (5 J *2 / 0.01 kg)1/2 = 31.6 m/s

Problems from Chap. 8

1. If gravity acts in the y-direction and an object is moving in the y-direction, is the momentum of the moving object constant (conserved)? Yes or no

No – gravity changes the velocity of the object, and p=mv.

2. A 1 kg object moving at 1 m/s runs into a stationary object, mass=4 kg. The collision is elastic (you must know what ‘elastic’ means when referring to a collision).

What is the velocity of the target (the 4 kg mass) after the collision? Remember from class 2/13 that when there are no external forces (→ momentum is conserved) and the collision is elastic (→ energy is conserved) that if mass m1 hits a stationary mass m2 that v1final = v1initial (m1-m2)/( m1+m2) and v2final = v1initial (2m1)/( m1+m2)

v2f = v1 initial * 2 * 1 kg/(1 kg + 4 kg) = 2/5 m/s

3. Two objects collide and stick together. What is this kind of collision called?

A ‘sticky’ one is NOT the right answer. When you read this, please email me at stewart@phys.ufl.edu and let me know a.) your answer, and b.) by sending the email I’ll know you got this far.

4. A 1 kg object, vinitial = 1 m/s, collides and sticks together with a 3 kg object that was initially stationary. What is the velocity after the collision of the two stuck together masses? units of m/s.

conserve momentum. pi = 1 kg * 1 m/s = pf = (1 kg + 3 kg) * vfinal

this implies that vfinal = ¼ m/s

5. In problem 4 just above, what fraction of the initial kinetic energy is lost in the collision? express as a percentage.

EKE = ½ mv2

EKEi = ½ 1 kg * (1 m/s)2 = ½ J. EKEf = ½ 4 kg * (1/4 m/s)2 = 1/8 J

so the fraction lost is (EKEi – EKEf)/ EKEi = ¾ = 75%

Problems from Chap. 9

1. A wheel with radius 10 cm is spinning at 1000 rpm. How fast (linear speed, in m/s) is the rim moving?

v=(r (=1000*2(/60 s = 104.7 rad/s so v=104.7 rad/s* 0.1 m = 10.5 m/s since the radius is 10 cm which is 0.1 of a meter. Note how the ‘unit’ of radians goes away – as discussed in class.

2. For #1 above, what is the angular velocity ( in rad/sec?

see above

3. For # 1 above, what is the angular acceleration ( in rad/s2?

the wheel is spinning with constant angular velocity. There is no change in the

velocity, ( is constant. Thus, (=0.

4. The tires of a car are 40 cm in outer radius (where they touch the road the tread of the tire is 40 cm from the center of the axle=axis of rotation.) If the car starts from rest and should reach 60 mi/hr in 6 seconds,

a. what angular velocity does the tire have at this 60 mi/hr speed, in rad/s?

if v=60 mi/hr (=26.82 m/s), we know this is equal to (r. Since r=0.4 m, this means that (=26.82 m/s /0.4 m = 67.1 rad/s. Note how the ‘unit’ radian appears at this point – you must have it for (, (, and ( whenever they have to do with the radius in an equation.

b. assuming constant angular acceleration (i. e. ( = (0 + (t, where ( is a constant), what is the linear acceleration of the edge of the tire (i. e. the outer part where the tread is 40 cm from the axis of rotation.) in m/s2.

well, we know that (0=0 (car starts from rest). We know from a.) that (=67.1 rad/s when the car has reached 60 mi/hr. We are told this acceleration took 6 seconds. Therefore, (=(( - (0)/t = 67.1 rad/s / 6 s = 11.2 rad/s2. Thus, the linear acceleration of the edge of the tire (which is the linear acceleration of the car itself – think about it!) is a=(r = 11.2 rad/s2 * 0.4 m = 4.47 m/s2. Note that this is almost ½ g of acceleration – you would be pressed back in your seat in this car while accelerating by a force almost ½ of your weight.

5. If some rotating object free from external torques and moving at angular speed 100 rad/s suddenly changes its moment of inertia I from Iinitial to twice the initial moment of inertia (i. e. Inew = 2 Iinitial), what is the new (?

Free from external torques means the angular momentum L = I( is constant. If I doubles, then ( must be halved. (new = ½ (beginning = 50 rad/s

6. A rotating object initially has 100 J of rotational kinetic energy. A torque is applied to increase the velocity of the rotating object until it now has 200 J of rotational kinetic energy. What is the new angular velocity in terms of the old one?

rotational K E = ½ I (2. If a torque is applied and the new rotational K E is doubled and since I, just like mass, is an intrinsic property that remains unchanged unless the problem says otherwise, then (2 must have doubled to double the rotational K E.

This implies that (new2 = 2 (old2. This implies that (new = (2)1/2 (old.

7. How much torque in Nm is needed to accelerate an object with I=100 kgm2 by 1 rad/s2?

torque = I( = 100 kgm2 * 1 rad/s2 = 100 kg m2/s2 = 100 Nm (since 1 N=1 kg m/s2)

8. What is the moment of inertia of a mass of 5 kg attached to the end of an essentially massless rod that is 7 m in lenth and which rotates around the center of the rod? (IN THE PICTURE, THE MASS on the top of the rod IS VERY Dense AND ESSENTIALLY HAS NO SIZE, WHILE EACH “X” is 1 m high. The arrow ‘→’ marks the point in the rod in the center where the axis of rotation is.)

M=5kg

X

X

X

→X

X

X

X

I=mr2, where r is the distance of the mass ‘m’ from the axis of rotation. Since this massless rod rotates around its center !, the right distance ‘r’ to use is 3.5 m. Thus, I=5 kg * (3.5 m)2 = 61.25 kgm2

Problems from Chap. 10 (only 1 lecture)

1. See the assigned problems, 11 and 13.

2. Consider the picture below. If the block weighs 980 N (=100 kg * 9.8 m/s2) and is in equilibrium, what are F1 and F2 in newtons, 3 sig figs?

Solution: Don’t need torques to do this, just need sum of forces in both the x and y directions to be separately = 0.

Fx = F1x + F2x

= - F1 cos30o + F2=0

Fy = F1y + F2y -980 N

= F1 sin 30o – 980 N

where of course F2y=0

Since both sums = 0, we know from the y-component equation that

F1 = 980 N/sin 30o so

F1 = 1960 N

this implies that

F2 = 1960 N*cos 30o

= 1697 N

Problems from Chap. 11 (1 lecture)

In addition to those ‘recommended’ (11-3,6,10), try:

1. If the density of Au (gold) is 19.3 g/cm3 and that of Pt (platinum) is 21.5 g/cm3, and someone offered to sell you 31.1 g of either (1 troy ounce), what would be the volumes involved of each?

31.1 g of Au takes up 1.611 cm3 (a cube 1.172 cm on a side, 0.462 inches)

31.1 g of Pt takes up 1.447 cm3 ( a cube 1.131 cm on a side, 0.445 inches)

2. How many grams of air are in someone’s lungs if the volume is 2000 cm3? (the density of air is about 0.0013 g/cm3)

every cm3 of the lungs contains 0.0013 g of air; if the lungs are 2000 cm3 in total volume, then total mass is 2000 cm3 * 0.0013 g/cm3 = 2.6 g.

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