AP Physics 1 Exam Question: Newton's Laws of Motion

Question: A crate of mass 20 kg is initially at rest on a horizontal floor. A 120 N horizontal force is applied to the crate, causing it to accelerate uniformly to the right. The coefficient of kinetic friction between the crate and the floor is 0.2. Calculate:

a) The magnitude of the force of kinetic friction acting on the crate. b) The acceleration of the crate. c) The velocity of the crate after it has traveled 5 meters.

Assume air resistance is negligible.

Answer:

a) To find the magnitude of the force of kinetic friction, we can use the formula:

Frictional force = Coefficient of friction * Normal force

The normal force is equal to the weight of the crate, so we can find the normal force using:

Normal force = Mass * Acceleration due to gravity

The weight of the crate is given by:

Weight = Mass * Acceleration due to gravity

Substituting these values, we have:

Normal force = 20 kg * 9.8 m/s^2 = 196 N
Frictional force = 0.2 * 196 N = 39.2 N

Therefore, the magnitude of the force of kinetic friction acting on the crate is 39.2 N.

b) Newton's second law states that the acceleration of an object is equal to the net force acting on it divided by its mass:

Acceleration = Net force / Mass

The net force acting on the crate is the applied force minus the force of kinetic friction:

Net force = Applied force - Frictional force = 120 N - 39.2 N = 80.8 N

Substituting these values, we have:

Acceleration = 80.8 N / 20 kg = 4.04 m/s^2

Therefore, the acceleration of the crate is 4.04 m/s^2.

c) We can use the equation for uniformly accelerated motion to find the final velocity of the crate:

Final velocity^2 = Initial velocity^2 + 2 * Acceleration * Distance

Since the crate starts from rest (initial velocity = 0 m/s), the equation simplifies to:

Final velocity^2 = 2 * Acceleration * Distance

Substituting the calculated values, we have:

Final velocity^2 = 2 * 4.04 m/s^2 * 5 m = 40.4 m^2/s^2

Taking the square root of both sides, we get:

Final velocity = √(40.4 m^2/s^2) = 6.35 m/s

Therefore, the velocity of the crate after it has traveled 5 meters is 6.35 m/s.