91Ó°ÊÓ

In this problem, there is no acceleration of the crate in the horizontal direction and the work done to move the crate across the floor is achieved by the pulling force.

The angle between the pushing force (F) and the direction of motion is zero.

Given data:

The weight of the crate is\(W = 1200\;{\rm{N}}\).

The distance is\(d = 5\;{\rm{m}}\).

The frictional force is\({F_{\rm{f}}} = 230\;{\rm{N}}\).

The free body diagram of the crate is as follows:

The relation between the forces in the x-direction is given by:

\(\begin{aligned}\Sigma {F_{\rm{x}}} &= 0\\F - {F_{\rm{f}}} &= 0\\F &= {F_{\rm{f}}}\end{aligned}\)

Here, F is the pushing force.

The relation between the forces in the x-direction is given by:

\(\begin{aligned}N - W &= 0\\N &= W &= 1200{\rm{ N}}\end{aligned}\)

The relation of work done is given by:

\({W_{{\rm{done}}}} = F \times d\cos {0^ \circ }\)

On plugging the values in the above relation, you get:

\(\begin{aligned}{W_{{\rm{done}}}} &= {F_{\rm{f}}} \times d\\{W_{{\rm{done}}}} &= \left( {230\;{\rm{N}}} \right)\left( {5\;{\rm{m}}} \right)\\{W_{{\rm{done}}}} &= 1150\;{\rm{J}}\end{aligned}\)

Thus, \({W_{{\rm{done}}}} = 1150\;{\rm{J}}\) is the required work done.

The relation of work done is given by:

\({W'_{{\rm{done}}}} = N \times d\cos {0^ \circ }\)

On plugging the values in the above relation, you get:

\(\begin{aligned}{{W'}_{{\rm{done}}}} &= \left( {1200\;{\rm{N}}} \right)\left( {5\;{\rm{m}}} \right)\\{{W'}_{{\rm{done}}}} &= 6000\;{\rm{J}}\end{aligned}\)

Thus, \({W'_{{\rm{done}}}} = 6000\;{\rm{J}}\) is the required work done.