The Physics
Hypertextbook
Opus in profectus

# Dynamics

## Practice

### practice problem 1

A person stands in an elevator weighing a cheeseburger with a kitchen scale. (It could happen.) The mass of the cheeseburger is 0.150 kg. The scale reads 1.14 N.
1. Draw a free body diagram showing all the forces acting on the cheeseburger.
2. Determine the weight of the cheeseburger.
3. Determine the magnitude and direction of the net force on the cheeseburger.
4. Determine the magnitude and direction of the elevator's acceleration.
5. At the time the person in the elevator is weighing the cheeseburger, the elevator's instantaneous velocity is upward. Is the speed of the elevator increasing, decreasing, or remaining constant at this moment? Justify your answer.

#### solution

Solutions…

1. All objects have weight. Objects resting on solid surfaces also experience a normal force. Weight points down, since it always does. Normal points up, since the problem didn't say anything about the scale not being level. Draw a box with one arrow pointing up and another pointing down. Try to make the upward pointing arrow look smaller than the downward one. Label the upward pointing arrow "normal" and the downward pointing arrow "weight".

2. Use the simple equation for weight. Assume the elevator is near the surface of the earth where gravity is around its standard value.

 W = mg W = (0.150 kg)(9.8 m/s2) W = 1.47 N
3. There are only two forces on the cheeseburger and they are opposite each other. This means the net force is the difference of the two forces. I think I will let up be the positive direction for this problem. The normal force is what the scale reads. Weight was computed in the previous part of this problem. The difference is negative, which means the net force is downward.

 ∑F = N − W ∑F = 1.14 N − 1.47 N ∑F = −0.33 N down
4. Use Newton's second law of motion to determine the acceleration. The mass of the cheeseburger was given in the problem and we just computed the net force a moment ago. Net force and acceleration are always in the same direction, since the math says so. Acceleration is also downward.

 a = ∑F m
 a = −0.33 N 0.150 kg
 a = −2.2 m/s2 down
5. The speed of the elevator is decreasing since the acceleration is opposite the velocity.

### practice problem 2

A 4.5 kg Canada goose is about to take flight. It starts from rest on the ground, but after a single step it is completely airborne. After 2.0 s of horizontal flight the bird has reached a speed of 6.0 m/s (fast enough to stay aloft, but not so fast that we need to worry about air resistance… at first).
1. Draw a free body diagram of the goose in flight.
2. Determine the following quantities for the goose in flight…
1. its acceleration
2. its weight
3. the magnitude and direction of the net force acting on it
4. the magnitude of the upward lift provided by its wings
5. the magnitude of the forward thrust provided by its wings
3. Any object moving through the air will experience air resistance. We just decided to ignore it temporarily. If we now admit that air resistance was present to some extent, how will this change the computed values of…
1. the acceleration?
2. the weight?
3. the net force?
4. the lift?
5. the thrust?
• All the measurements given in the problem are still valid for part c of this problem. The mass is still 4.5 kg and the bird still accelerates from rest to 6.0 m/s in 2.0 s.

#### solution

The solutions…

1. All objects have weight. It points down. A winged object like a bird needs some force to keep it aloft. In aerodynamics, they call it lift. An object accelerating forward must have some force pushing it forward. In aerodynamics, they call it thrust. If there was drag, it would point opposite the direction of the bird's motion (in other words, backward). 2. Determine the following quantities for the goose in flight…

1. In this problem, acceleration is computed from its definition.

 a = ∆v ∆t
 a = 6.0 m/s − 0 m/s 2.0 s
 a = 3.0 m/s2 forward
2. Weight is mass times gravity (the value of the gravitational field on earth). Weight always points down. (Down is defined as the direction things move when allowed to fall freely.)

 W = mgW = (4.5 kg)(9.8 m/s2)W = 44.1 N down
3. Use Newton's second law of motion. Net force and acceleration always have the same direction.

 ∑F = ma∑F = (4.5 kg)(3.0 m/s2)∑F = 13.5 N forward
4. There is no net force in the vertical direction. Thus, whatever forces act up must be balanced by forces acting down. According to our diagram, this means lift equals weight.

L = −W = 44.1 N up

5. The forward force of thrust is the only force acting horizontally, which makes it the net force.

T = ∑F = 13.5 N forward

3. What if air resistance was not negligible? How would this change the values calculated above? How would this change the…

1. Acceleration was computed from measured values of velocity and time. Adding drag to the problem does not change these measurements, therefore acceleration does not change.
2. Weight is determined by mass and gravity. Mass is an invariant quantity. Nothing can change it. The value of the gravitational field is determined by the location. Adding drag does not change the mass of the bird or its location, therefore weight does not change.
3. The net force is determined by mass and acceleration. Neither of these quantities is affected by drag, therefore net force does not change.
4. Lift balances weight. Since weight does not change, lift does not change.
5. When there was no drag, the thrust accelerated the bird. Now that there is drag, so the thrust has to accelerate the bird and overcome drag. To maintain the same acceleration, thrust would have to increase.

### practice problem 3

A laboratory cart (m1 = 500 g) is pulled horizontally across a level track by a lead weight (m2 = 25 g) suspended vertically off the end of a pulley as shown in the diagram below. (Assume the string and pulley contribute negligible mass to the system and that friction is kept low enough to be ignored.) 1. Draw a free body diagram for…
1. the cart
2. the weight
2. Determine…
1. the acceleration of the system
2. the tension in the string