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olga55 [171]
3 years ago
7

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 di

splacement is 65 m. What is the total work done on the crate by the frictional force?
Physics
1 answer:
trapecia [35]3 years ago
8 0

Answer:

The work done by the friction force is 11700J

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

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