Question

The truck is accelerating at a rate of +1.50 m/s2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by the frictional force?

Answer 1

The work done by the friction force is 11700J

Step-by-step explanation:

The work done by the friction force is given by:

`Wfs=fs*cos(\theta)*d\\Wfs=fs*cos(0)*65m`

According to Newton's second law:

`f=m*a\\fs=120kg*1.50m/s^2=180N`

So:

`Wfs=180N*65m\\Wfs=11700J`