AP Physics 1 Exam Question

A wooden block hangs from a string that is attached to the ceiling. The mass of the block is 2 kg. A horizontal force of 10 N is applied to the block, causing it to accelerate to the right.

1. Determine the tension in the string while the block is accelerating.

2. After some time, the block reaches a constant velocity due to air resistance. Find the magnitude and direction of the normal force acting on the block at this point.

Assume air resistance is the only external force acting on the block, with a magnitude of 2 N. Take the acceleration due to gravity as 9.8 m/s².

Answer:

1. To determine the tension in the string while the block is accelerating, we need to find the net force acting on the block.

The applied force, denoted by F_app, is responsible for accelerating the block while pulling it to the right. The weight of the block, mg, acts vertically downward with a magnitude of 2 kg × 9.8 m/s² = 19.6 N.

The net force, F_net, acting on the block can be calculated using Newton's second law: F_net = ma

Where:

• F_net is the net force,
• m is the mass of the block,
• and a is the acceleration of the block.

In this case, the net force is equal to the applied force (F_app), as air resistance is ignored.

Therefore, F_app = ma.

Rearranging the equation to solve for tension (T):

T = F_net = F_app = ma

T = (2 kg) × (10 m/s²)

T = 20 N

Hence, the tension in the string while the block is accelerating is 20 N.

2. To find the magnitude and direction of the normal force when the block reaches a constant velocity, we need to consider the forces acting on the block.

At a constant velocity, the air resistance force must be balanced by the tension in the string and the block's weight.

The weight of the block, mg, is still vertically downward with a magnitude of 19.6 N.

The air resistance force, Fr, is equal to 2 N, acting opposite to the direction of motion.

The normal force, N, acts perpendicular to the contact surface (ceiling in this case) and keeps the block suspended.

Since the block is not accelerating at constant velocity, the net force must be zero:

F_net = ma = 0

Taking upward as positive and downward as negative, the forces can be summed up:

N - mg - Fr = 0

Simplifying the equation:

N = mg + Fr

N = 19.6 N + 2 N

N = 21.6 N

Therefore, the magnitude of the normal force acting on the block at this point is 21.6 N. The normal force is directed upward, opposing the downward weight of the block.

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