The Physics
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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) rests on a level track. It is connected to a lead weight (m2 = 100 g) suspended vertically off the end of a pulley as shown in the diagram below. The system is released and the cart accelerates to the right. (Assume the string and pulley contribute negligible mass to the system and that friction is kept low enough that it can be ignored.) Draw a free body diagram for…
1. the laboratory cart
2. the lead weight
Determine…
1. the weight of the lead weight in newtons
2. the weight of the laboratory cart in newtons
3. the normal force of the cart on the track
4. the net force acting on the system
5. the acceleration of the system
6. the tension in the string

#### solution

The challenge in this problem is keeping track of the different objects. Sometimes we're dealing with the lab cart (identified by a subscripted 1), sometimes we're dealing with the lead weight (identified by a subscripted 2), and sometimes we're dealing with the whole system — the cart and weight connected by a string (identified by the lack of a subscript). This level of detail is not necessary for your own personal work, but it is a good idea for me to do it so that my work is less ambiguous to you.

1. Why make two diagrams when you can make one?

2. Lab cart on the left, lead weight on the right. 3. Weight is mass times gravity. The SI unit of force is the newton, which is based on the kilogram and the meter per second squared. Be sure you're using the right units.

 W2 = m1gW2 = (0.100 kg)(9.8 m/s2)W2 = 0.980 N
4. Repeat the steps above with a different mass.

 W1 = m1gW1 = (0.500 kg)(9.8 m/s2)W1 = 4.90 N
5. Normal equals weight on a level surface like the one described above.

N1 = W1
N1 = 4.90 N

6. Normal and weight cancel out on the track. Tension is an internal force for the cart-weight system. The net force is whatever's left over — the weight of the lead weight.

F = W2
F = 0.980 N

7. Use Newton's second law of motion to determine the acceleration of the system. The mass that's being accelerated is the mass of the cart plus the weight.

 a = ∑F/ma = (0.980 N)/(0.500 kg + 0.100 kg)a = 1.63 m/s2

Note that this is less than the acceleration due to gravity, which is as it should be. The system is not in free fall.

8. Tension is an internal force for the system as a whole, but it is the net force acting on the cart. There is nothing balancing it. Apply Newton's second law to the cart by itself. (Let right be the positive direction since that's the direction the lab cart is accelerating.)

 ∑F1 = m1a∑T = (0.500 kg)(1.63 m/s2)∑T = 0.816 N right

Tension is also one of two forces acting on the suspended weight. The other is the weight of the weight. The difference in these two is the net force on the lead weight. Use this information and Newton's second law to find the tension. (Let down be the positive direction since that's the direction the lead weight is accelerating.)

 ∑F2 = m2a W2 − T = m2a (0.980 N) − T = (0.100 kg)(1.63 m/s2) T = 0.816 N up

Two methods give the same answer, so all is well. The lab cart and the lead weight experience the same tension (same magnitude, different directions).

### practice problem 4

Write something completely different.

Answer it.