# Unit 3 Concepts Quiz

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• 1.

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

• A.

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

• B.

There are no forces acting on it at all

• C.

The net force on the book is zero

• D.

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

C. The net force on the book is zero
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.

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• 2.

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

• A.

More than its weight

• B.

Equal to its weight

• C.

Less than its weight but more than zero

• D.

Depends on the speed of the puck

• E.

Zero

E. Zero
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.

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• 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?

• A.

Slowly come to a stop

• B.

Continue with constant acceleration

• C.

Continue with constant velocity

• D.

Continue with decreasing acceleration

• E.

Immediately come to a stop

C. Continue with constant velocity
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.

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• 4.

### 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?

• A.

4 v

• B.

2 v

• C.

V

• D.

1/2 v

• E.

1/4 v

D. 1/2 v
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.

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• 5.

### 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?

• A.

Fg is greater on the feather

• B.

Fg is greater on the stone

• C.

Fg is zero on both due to vacuum

• D.

Fg is equal on both always

• E.

Fg is zero on both always

B. Fg is greater on the stone
Explanation
The force of gravity (weight) depends on the mass of the object!! The stone has more mass, therefore more weight.

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• 6.

### 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?

• A.

G is greater on the feather

• B.

G is greater on the stone

• C.

G is zero on both due to vacuum

• D.

G is equal on both always

D. G is equal on both always
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.

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• 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?

• A.

N > mg

• B.

N = mg

• C.

N < mg (but not zero)

• D.

N = 0

• E.

Depends on the size of the elevator

B. N = mg
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.

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• 8.

### 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?

• A.

N > mg

• B.

N = mg

• C.

N < mg (but not zero)

• D.

N = 0

• E.

Depends on the size of the elevator

A. N > mg
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!

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• 9.

### 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?

• A.

Block on flat surface

• B.

Block on incline

• C.

Both the same (N = mg)

• D.

Both the same (N < mg)

• E.

Both the same (N = 0)

A. Block on flat surface
Explanation
for the flat surface, we know that N = W. For the incline, due to the angle of the incline, N < W. In fact, we can see that N = W cos(θ). Since Cos(θ) can only be between 0 and 1 we know N must be less than W.

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• 10.

### 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??

• A.

• B.

You don’t go up, you’re too heavy

• C.

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

• D.

The rope actually pulls you up

• E.

You are pulling the ceiling down

D. The rope actually pulls you up
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.

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• 11.

### 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?

• A.

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

• B.

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

• C.

The forces are equal

• D.

The forces are zero because they cancel out

• E.

There are actually no forces at all

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

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• 12.

### 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?

• A.

They do not accelerate because they are weightless

• B.

Accelerations are equal, but not opposite

• C.

Accelerations are opposite, but bigger for the bowling ball

• D.

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

• E.

Accelerations are equal and opposite

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

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• 13.

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

• A.

Moves to the left

• B.

Moves to the right

• C.

Moves up

• D.

Moves down

• E.

The box does not move

E. The box does not move
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.

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• 14.

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

• A.

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

• B.

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

• C.

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

• D.

Zero if the rate of rotation is constant

• E.

Independent of the person's mass

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

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• 15.

### 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).

• A.

4000 N

• B.

14000 N

• C.

6000 N

• D.

10000 N

B. 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.

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• 16.

• A.

4 N

• B.

100 N

• C.

20 N

• D.

16 N