# Questions (review for final)

1. Dec 19, 2003

### moonlit

Hi,

I'm reviewing for my final and was wondering if someone can help me out with some problems I've gotten wrong on previous tests. I was wondering what the correct answers (or atleast correct formulas) would be used to solve these problems. Thanks.

1) A 74.0 kg person stands on a scale in an elevator. Find the apparent weight, (ex. the scale reading) for case a and case b.
a) the elevator is going up with a constant velocity of 8.11 m/s.
b) the elevator is going down with an acceleration of 1.80 m/s^2

2) A 95 kg baseball slides into home plate. The coefficient of kinetic friction between the player and the ground is .45
a) what is the magnitude of the frictional force?
b) if the player goes into the slid with a speed of 6.4 m/s. How long will it take him to come to rest (in seconds)?

3) A man tosses a bottle of water with a mass of 0.36 kg up to a life guard. She catches it 2.8 m above the ground. The water bottle leaves the man's hand 0.6 m above the ground.
a) Find the potential energy of the water bottle relative to the ground when it's caught.
b) Find the potential energy of the water bottle relative to the ground when it leaves the man's hand.
c) How much work is done on the water bottle by the force of gravity during the throw?

4) The combined mass of a car and driver is 1800 kg. If the road is dry and flat the maximum frictional force between the tires and the road is 15000 N. Calculate the minimum radius of a curve the car can safely negotiate at 23.0 m/s.

5) A person can push a lawn mower with constant velocity for a distance of 26.0 m across the lawn. He pushes on the handle at an angle of 35 degrees below the horizontal with a force of 100 N.
a) Find the work done on the lawn mower by the force applied to the handle.
b) How much work is done by the force of friction.

6) A stray golf ball comes straight down and lands in the parking lot. The mass of the golf ball is .045 kg and its speed is 38.0 m/s just before and after it hits the pavement.
a) Find the magnitude of the impulse applied to the golf ball by the pavement.
b) If the golf ball is in contact with the pavement for 11.0 ms find the average force applied to the golf ball.

7) A skateboarder is coasting along a level sidewalk with a speed of 2.70 m/s. The skater's mass is 52.0 kg. His backpack is sitting on the ground and as he glides past it he quickly grabs it and continues. Find the speed of the skateboarder after he grabs the backpack. The mass of the backpack is 7.0 kg. (Friction is negligible and the skateboarder does not push or kick to power himself.)

8) The angular speed of a rotor in a centrifuge increases from 220 rad/s to 1180 rad/sec in 4.00 seconds.
a) Find the angle through which the rotor turns in that time.
b) Find the magnitude of the angular acceleration.

9) A clay vase on a potter's wheel goes through an angular acceleration of 2.1 rev/s^2 due to the application of a torque of 12.0 Nxm. Find the moment of inertia of the combined vase and potter's wheel.

2. Dec 19, 2003

### jamesrc

OK, that's a lot of questions. I'll start typing as many explanations as I have time to, and if I don't get to all of them, I'm sure someone else will. Please include your work/ideas on these problems if you want any further clarification.

1. Draw a free body diagram. When the elevator moves with constant velocity, there is no net force on the person. The net force on the person is N-mg (taking up to be positive). ma = N - mg (Newton's 2nd law), so for constant velocity, a = 0 and N = mg. (N, the normal force is the scale reading in this problem). For the other part, plug in a = -1.8 m/s/s and solve for N.

2. A 95 kg baseball? Oh, you mean baseball player. For part a, the kinetic friction is given by f = &mu;kN, where the normal force N = mg. For the second part, you have a constant deceleration (given by ma = -f) and an initial velocity. You should know the kinematic equations for constant acceleration like the back of your hand by now; use the one that fits the given information and allows you to solve for time.

3. Gravitational potential is given by U = mgh, where h is the height off of the ground. In general, you get to decide where "ground" is, but it's implied in this problem that THE ground is ground (that's not nearly as convoluted as it sounds). To find the work done by the gravitational force, you just need to multiply the force by the distance. The work done by the force of gravity on the bottle is negative here since the displacement vector and the force vector are antiparallel.

4. Equate the centripetal force with the frictional force.

5. Since it is moving at a constant velocity, there is no net work done on the mower (work by pushing + work by friction = 0). Resolve the 100N force into components perpendicular and parallel to the ground. For a constant applied force, the work done by that force is simply the dot product of force and displacement.

6. Impulse is equal to change in momentum. I = mv - mvo. Given a time for the impulse the average force is given by I = F&Delta;t

3. Dec 19, 2003

### jamesrc

Finishing up:

7. Conservation of momentum. Mv = (M+m)v'

8. Assuming constant angular acceleration,
$\alpha = \frac{\Delta \omega}{\Delta t}$
To find the angle that it rotates through, remember that there are relations for rotational kinematics that are analagous to the translational ones.

9. Use the rotational analog of Newton's second law:
$\vec\tau_{\rm net} = I\vec{\alpha}$