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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?

Respuesta :

krlosdark12

Answer:

The work done by the friction force is 11700J

Explanation:

The work done by the friction force is given by:

[tex]Wfs=fs*cos(\theta)*d\\Wfs=fs*cos(0)*65m[/tex]

According to Newton's second law:

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

So:

[tex]Wfs=180N*65m\\Wfs=11700J[/tex]

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