AP Physics Unit 2 review "flashcard"

Front

The sum of the forces on the object is zero in which of the cases? Options: II only, III only, I and II only, I and III only.

Back

I and II only

Answer explanation

The sum of the forces is zero when the objects are at constant velocity or at rest. In graphs I and II, the position is linear, indicating constant velocity. In graph III, the position is accelerating, so forces are not balanced.

Front

Two people are pulling on the ends of a rope. Each person pulls with a force of 100 N. The tension in the rope is:

Back

100 N

Answer explanation

The tension in the rope is equal to the force exerted by each person pulling on it. Since both are pulling with a force of 100 N, the tension in the rope is 100 N.

Front

A student pulls a wooden box along a rough horizontal floor at constant speed by means of a force P. Which of the following must be true?

Back

P > f and N < W.

Answer explanation

Since the box is pulled at constant speed, the pulling force P must be greater than the frictional force f (P > f). The normal force N is less than the weight W because some of the weight is countered by the vertical component of P, thus N < W.

Front

A tractor-trailer truck is traveling down the road. The mass of the trailer is 4 times the mass of the tractor. If the tractor accelerates forward, what is the force that the trailer applies on the tractor?

Back

equal to the force of the tractor on the trailer.

Answer explanation

According to Newton's third law, the force exerted by the tractor on the trailer is equal in magnitude and opposite in direction to the force exerted by the trailer on the tractor. Thus, they are equal.

Front

A 500-gram ball moving at 15 m/s slows down uniformly until it stops. If the ball travels 15 meters, what was the average net force applied while it was coming to a stop?

Back

3.75 newtons

Answer explanation

To find the average net force, use F = ma. First, calculate the deceleration: v^2 = u^2 + 2as. Here, u = 15 m/s, v = 0, and s = 15 m. Solving gives a = -1.25 m/s². Then, F = 0.5 kg * -1.25 m/s² = -0.625 N. The average net force is 3.75 N.

Front

A homeowner pushes a lawn mower across a horizontal patch of grass with a constant speed by applying a force P. Which of the following relations among the various force magnitudes, W, f, N, P is CORRECT? Options: P > f and N > W, P < f and N = W, P > f and N < W, P = f and N > W

Back

P > f and N > W

Answer explanation

The homeowner applies a force P to move the mower at constant speed, meaning P must be greater than the friction force f (P > f). The normal force N must also be greater than the weight W of the mower (N > W) due to the vertical forces acting.

Front

An object originally traveling at a velocity, v₀, is accelerated to a velocity, v, in a time, t, by a constant force, F. What would be the mass of the object?

Back

Answer explanation

Using Newton's second law, F = ma, we can express mass as m = F/a. The acceleration a can be calculated as (v - v₀)/t. Substituting this into the mass equation gives m = Ft/(v - v₀), which matches the correct answer.

Front

A tennis ball is thrown against a vertical concrete wall and bounces off. How does the force exerted by the ball on the wall compare with the force exerted by the wall on the ball?

Back

The forces exerted by the ball and the wall have the same magnitude.

Answer explanation

According to Newton's third law, the forces exerted by the ball on the wall and the wall on the ball are equal in magnitude and opposite in direction. Therefore, the correct choice is that the forces have the same magnitude.

Front

The diagram above represents the forces exerted on a box that a child is holding. F_N represents the force applied by the child’s hand, and F_g represents the weight of the box. The child begins to raise the box with increasing speed. Which of the following claims is correct about force F_h that is exerted by the box on the child’s hand as the box is being raised? Options: F_h=F_N, where F_N does not change as the child raises the box.; F_h=F_N, where F_N is larger as the box is being raised than when it was being held.; F_h=F_g, where F_g does not change as the child raises the box.; F_h=F_g, where F_g is larger as the box is being raised than when it was being held.

Back

F_h=F_N, where F_N is larger as the box is being raised than when it was being held.

Answer explanation

As the child raises the box with increasing speed, the force F_N exerted by the child's hand must increase to overcome both the weight F_g and provide the upward acceleration. Thus, F_h = F_N, where F_N is larger when raising the box.

Front

By doubling both the hanging mass and the cart mass in a Newton's laws experiment, how will the tension in the string and the acceleration of the cart change?

Back

The tension will double, but the acceleration will stay the same.

Answer explanation

Doubling both masses keeps the ratio of forces the same, so the acceleration remains constant. However, the tension in the string increases due to the greater weight of the hanging mass, thus it will double.