Question

A man pushing a crate of mass m = 92.0 kg at a speed of  v = 0.880 m/s encounters a rough horizontal surface of length ℓ = 0.65 m as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.357 and he exerts a constant horizontal force of 297 N on the crate.

A man pushes a crate labeled m, which moves with a velocity vector v to the right, on a horizontal surface. The horizontal surface is textured from the right edge of the crate to a horizontal distance ℓ from the right edge of the crate.

(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.

magnitude	N
direction	---Select--- opposite as the motion of the crate same as the motion of the crate

(b) Find the net work done on the crate while it is on the rough surface.
J

(c) Find the speed of the crate when it reaches the end of the rough surface.
m/s

Answer 1

Part A

Magnitude of the net force will be the difference between the fricitonal force and the applied force.

We know the applied force already. The frictional force is the weight of the crate times the coefficient of friction.

Therefore, the net force is

The negative sign indicates that the force is in the direction opposite to the motion of the crate.

The magnitude of the net force is simply the modulus of this.

Part B

To find the work done by the net force, we multiply it by the distance over which the force acts.

The negative sign indicates that the crate loses velocity as it travels over the rough surface.

Part C

There are two ways to calculate the speed of the crate after it travels over the rough surface. The first method is to find the deceleration of the crate, and use the equations of kinematics to find the final speed. The other method is to find the final kinetic energy of the crate after the friction does negative work on it.

• First method.

First, we find the acceleration

We know that

• Second method