1.

A mover pushes a large crate (mass m=75 kg) from the inside of the truck to the back end (a distance of 6 m), exerting a steady push of 300 N. if he moves the crate this distance in 20 s, what is his power output during this time?

Answer»

SOLUTION :The work done on the CRATE by the mover is `W=Fd=(300N)(6m)=1,800J`. If this much work is done is 20 s, then the POWER delivered is `P=W//t=(1,800J)//(20s)`
=90Wgt Note that `P=W//t=Fd//t=Fv`, the formula P=Fv is often useful.


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