Question

Derek pushes a 10kg crate along a rough surface with 35N of force. He observes that it is only accelerating at a rate of 2ms2. What must the value of the force of friction be? (Assume the only two forces acting on the object are friction and Derek).

a) 5 N
b) 15 N
c) 25 N
d) 45 N

Answer 1

The force of friction experienced by the system consisting of Derek and the crate is 55 N.

Step-by-step explanation:

In order to find the force of friction experienced by the system, we need to calculate the net force acting on the crate.

According to Newton's second law of motion, the net force on an object is equal to the mass of the object multiplied by its acceleration:

Fnet = m * a

Substituting the given values, we have:

Fnet = 10 kg * 2 m/s2 = 20 N

The only two forces acting on the crate are the force applied by Derek (35 N) and the force of friction. Since the crate is accelerating, the force of friction must be in the opposite direction of the applied force.

Therefore, the force of friction is:

Ffriction = -35 N - 20 N = -55 N

Since the force of friction is a negative value, the magnitude of the force of friction is 55 N. However, the question asks for the value of the force of friction, so the correct answer is 55 N.