a worker pushes a 1500N crate with a horizontal force of 345Na distance of 24 me

Question

a worker pushes a 1500N crate with a horizontal force of 345Na distance of 24 meters. Assume the coefficient of kinetic frictionbetween the crate and the floor is 0.220. a. How much work is done by the worker on the crate? b.How much work is done by the floor on the crate? c.what is the net work done on the crate? a worker pushes a 1500N crate with a horizontal force of 345Na distance of 24 meters. Assume the coefficient of kinetic frictionbetween the crate and the floor is 0.220. a. How much work is done by the worker on the crate? b.How much work is done by the floor on the crate? c.what is the net work done on the crate?

Explanation / Answer

(a) The work done by worker on the crate                         W1 = F*s = 345 N * 24m = 8280 J (b) The work done by floor on the crate                         W2 = mg *s = (0.22) (1500) (24m)                               = 7920 J (c) the net work done   W = W1 - W2 = 8280 J- 7920 J = 360 J

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