Unit 3 Concepts Quiz

1. A feather and a stone fall together in a vacuum tube with no air resistance. What can you say about the acceleration of gravity "g" acting on the stone and the feather?

G is greater on the feather

G is greater on the stone

G is zero on both due to vacuum

G is equal on both always

In a vacuum tube with no air resistance, the acceleration of gravity "g" is the same for both the feather and the stone. This is because the acceleration due to gravity is solely determined by the mass of the object and the gravitational force acting on it, and is independent of the object's size or shape. Therefore, the feather and the stone will experience the same acceleration of gravity "g" as they fall together in the vacuum tube.

Explanation

In a vacuum tube with no air resistance, the acceleration of gravity "g" is the same for both the feather and the stone. This is because the acceleration due to gravity is solely determined by the mass of the object and the gravitational force acting on it, and is independent of the object's size or shape. Therefore, the feather and the stone will experience the same acceleration of gravity "g" as they fall together in the vacuum tube.

2. A book is lying at rest on a table. The book will remain there at rest because:

There is a net force but the book has too much inertia

There are no forces acting on it at all

The net force on the book is zero

There is a net force, but the book is too heavy to move

There are forces acting on the book, but the only forces acting are in the y-direction. Gravity acts downward, but the table exerts an upward force that is equally strong, so the two forces cancel, leaving no net force.

Explanation

There are forces acting on the book, but the only forces acting are in the y-direction. Gravity acts downward, but the table exerts an upward force that is equally strong, so the two forces cancel, leaving no net force.

3. Consider a cart on a horizontal frictionless table. Once the cart has been given a push and released, what will happen to the cart?

Slowly come to a stop

Continue with constant acceleration

Continue with constant velocity

Continue with decreasing acceleration

Immediately come to a stop

After the cart is released, there is no longer a force in the x-direction. This does not mean that the cart stops moving!! It simply means that the cart will continue moving with the same velocity it had at the moment of release. The initial push got the cart moving, but that force is not needed to keep the cart in motion.

Explanation

After the cart is released, there is no longer a force in the x-direction. This does not mean that the cart stops moving!! It simply means that the cart will continue moving with the same velocity it had at the moment of release. The initial push got the cart moving, but that force is not needed to keep the cart in motion.

4. A hockey puck slides on ice at constant velocity. What is the net force acting on the puck?

More than its weight

Equal to its weight

Less than its weight but more than zero

Depends on the speed of the puck

Zero

The puck is moving at a constant velocity, and therefore it is not accelerating. Thus, there must be no net force acting on the puck.

Explanation

The puck is moving at a constant velocity, and therefore it is not accelerating. Thus, there must be no net force acting on the puck.

5. When you climb up a rope, the first thing you do is pull down on the rope. How do you manage to go up the rope by doing that??

This slows your initial velocity which is already upward

You don’t go up, you’re too heavy

You’re not really pulling down – it just seems that way

The rope actually pulls you up

You are pulling the ceiling down

When you pull down on the rope, the rope pulls up on you!! It is actually this upward force by the rope that makes you move up! This is the “reaction” force (by the rope on you) to the force that you exerted on the rope. And voilá, this is Newton’s 3rd Law.

Explanation

When you pull down on the rope, the rope pulls up on you!! It is actually this upward force by the rope that makes you move up! This is the “reaction” force (by the rope on you) to the force that you exerted on the rope. And voilá, this is Newton’s 3rd Law.

6. A force F acts on mass M for a time interval T, giving it a final speed v. If the same force acts for the same time on a different mass 2M, what would be the final speed of the bigger mass?

4 v

2 v

V

1/2 v

1/4 v

In the first case, the acceleration acts over time T to give velocity v = aT. In the second case, the mass is doubled, so the acceleration is cut in half, therefore, in the same time T, the final speed will only be half as much.

Explanation

In the first case, the acceleration acts over time T to give velocity v = aT. In the second case, the mass is doubled, so the acceleration is cut in half, therefore, in the same time T, the final speed will only be half as much.

7. A block of mass m rests on the floor of an elevator that is moving upward at constant speed. What is the relationship between the force due to gravity and the normal force on the block?

N > mg

N = mg

N < mg (but not zero)

N = 0

Depends on the size of the elevator

The block is moving at constant speed, so it must have no net force on it. The forces on it are N (up) and mg (down), so N = mg, just like the block at rest on a table.

Explanation

The block is moving at constant speed, so it must have no net force on it. The forces on it are N (up) and mg (down), so N = mg, just like the block at rest on a table.

8. When a person stands on a rotating merry-go-round, the frictional force exerted on the person by the merry-go-round is:

Greater in magnitude than the frictional force exerted on the person by the merry-go-round

Opposite in direction to the frictional force exerted on the merry-go-round by the person

Directed away from the center of the merry-go-round

Zero if the rate of rotation is constant

Independent of the person's mass

When a person stands on a rotating merry-go-round, the frictional force exerted on the person by the merry-go-round is opposite in direction to the frictional force exerted on the merry-go-round by the person. This is because of Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. As the person pushes against the merry-go-round, the merry-go-round pushes back with an equal force in the opposite direction. This frictional force between the person and the merry-go-round helps to keep the person in place while the merry-go-round rotates. The magnitude of the frictional force depends on factors such as the coefficient of friction and the normal force, but it is not greater in magnitude than the frictional force exerted by the person. The direction of the frictional force is opposite to the force exerted by the person, not directed away from the center of the merry-go-round. The frictional force is not zero if the rate of rotation is constant, and it is also independent of the person's mass.

Explanation

When a person stands on a rotating merry-go-round, the frictional force exerted on the person by the merry-go-round is opposite in direction to the frictional force exerted on the merry-go-round by the person. This is because of Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. As the person pushes against the merry-go-round, the merry-go-round pushes back with an equal force in the opposite direction. This frictional force between the person and the merry-go-round helps to keep the person in place while the merry-go-round rotates. The magnitude of the frictional force depends on factors such as the coefficient of friction and the normal force, but it is not greater in magnitude than the frictional force exerted by the person. The direction of the frictional force is opposite to the force exerted by the person, not directed away from the center of the merry-go-round. The frictional force is not zero if the rate of rotation is constant, and it is also independent of the person's mass.

9. A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block?

N > mg

N = mg

N < mg (but not zero)

N = 0

Depends on the size of the elevator

The block is accelerating upward, so it must have a net upward force. The forces on it are N (up) and mg (down), so N must be greater than mg in order to give the net upward force!

Explanation

The block is accelerating upward, so it must have a net upward force. The forces on it are N (up) and mg (down), so N must be greater than mg in order to give the net upward force!

10. A box of weight 100 N is at rest on a floor where μ_s = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?

Moves to the left

Moves to the right

Moves up

Moves down

The box does not move

The static friction force has a maximum of μsN = 40 N. The tension in the rope is only 30 N. So the pulling force is not big enough to overcome friction.

Explanation

The static friction force has a maximum of μsN = 40 N. The tension in the rope is only 30 N. So the pulling force is not big enough to overcome friction.

11. Consider two identical blocks, one resting on a flat surface, and the other resting on an incline. For which case is the normal force greater?

Block on flat surface

Block on incline

Both the same (N = mg)

Both the same (N < mg)

Both the same (N = 0)

for the flat surface, we know that N = W. For the incline, due to the angle of the incline, N

Explanation

for the flat surface, we know that N = W. For the incline, due to the angle of the incline, N

12. In outer space, a bowling ball and a ping-pong ball attract each other due to gravitational forces. How do the magnitudes of these attractive forces compare?

The bowling ball exerts a greater force on the ping-pong ball

The ping-pong ball exerts a greater force on the bowling ball

The forces are equal

The forces are zero because they cancel out

There are actually no forces at all

The forces are equal and opposite by Newton’s 3rd Law!

Explanation

The forces are equal and opposite by Newton’s 3rd Law!

13. A 5-kg block is being pushed by a 20-N horizontal force so that the block is moving at a constant velocity of 0.8 m/s along a level floor. The frictional force acting on the block by the floor is:

4 N

100 N

20 N

16 N

If it is moving at a constant velocity then the net force in the x-direction is zero. Therefore Force of friction = force of the pull (20 N).

Explanation

If it is moving at a constant velocity then the net force in the x-direction is zero. Therefore Force of friction = force of the pull (20 N).

14. A rocket with a mass of 1000 kg is accelerating upward. The acceleration is 4 m/s² upward. What is the magnitude of the thrust (lifting force) of the rocket's engines? (use g = 10 m/s²).

4000 N

14000 N

6000 N

10000 N

The thrust (lifting force) of the rocket's engines can be calculated using Newton's second law of motion, which states that force equals mass multiplied by acceleration. In this case, the mass of the rocket is 1000 kg and the acceleration is 4 m/s2 upward. Therefore, the thrust can be calculated as 1000 kg * 4 m/s2 = 4000 N. However, it is important to note that the question mentions using g = 10 m/s2, which is the acceleration due to gravity. This means that the upward acceleration of the rocket is actually the result of the engines overcoming the force of gravity, so the thrust of the engines must be greater than the weight of the rocket. Therefore, the correct answer is 14000 N.

Explanation

The thrust (lifting force) of the rocket's engines can be calculated using Newton's second law of motion, which states that force equals mass multiplied by acceleration. In this case, the mass of the rocket is 1000 kg and the acceleration is 4 m/s2 upward. Therefore, the thrust can be calculated as 1000 kg * 4 m/s2 = 4000 N. However, it is important to note that the question mentions using g = 10 m/s2, which is the acceleration due to gravity. This means that the upward acceleration of the rocket is actually the result of the engines overcoming the force of gravity, so the thrust of the engines must be greater than the weight of the rocket. Therefore, the correct answer is 14000 N.

15. A feather and a stone fall together in a vacuum tube with no air resistance. What can you say about the force of gravity Fg acting on the stone and the feather?

Fg is greater on the feather

Fg is greater on the stone

Fg is zero on both due to vacuum

Fg is equal on both always

Fg is zero on both always

The force of gravity (weight) depends on the mass of the object!! The stone has more mass, therefore more weight.

Explanation

The force of gravity (weight) depends on the mass of the object!! The stone has more mass, therefore more weight.

16. In outer space, gravitational forces exerted by a bowling ball and a ping-pong ball on each other are equal and opposite. How do their accelerations compare?

They do not accelerate because they are weightless

Accelerations are equal, but not opposite

Accelerations are opposite, but bigger for the bowling ball

Accelerations are opposite, but bigger for the ping-pong ball

Accelerations are equal and opposite

The forces are equal and opposite -- this is Newton’s 3rd Law!! But the acceleration is F/m and so the smaller mass has the bigger acceleration.

Explanation

The forces are equal and opposite -- this is Newton’s 3rd Law!! But the acceleration is F/m and so the smaller mass has the bigger acceleration.