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A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220.

AHow much work is done by the worker on the crate?
BHow much work is fone by the floor on the crate?
CWhat is the net work done on the crate?

Verified Answer :

onyebuchinnaji

(a) The work done by the worker on the crate is 8,280 J.

(b) The work done by the floor on the crate is 7,920 J.

(c) The net work done on the crate is 360 J.

Work done by the worker

The total work done by the worker is calculated as follows;

\(W = Fd\\\\W = 345 \times 24\\\\W = 8,280 \ J\)

Work done by the floor on the crate

The work done by the floor on the crate is calculated as follows;

\(W = F_f d\\\\W = \mu F_n d\\\\W = 0.22 \times 1.5 \times 10^3 \times 24\\\\W = 7,920 \ J\)

Net work done on the crate

The net work done on the crate is calculated as follows;

\(W = 8,280 - 7,920\\\\W = 360 \ J\)

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