Question

(II) How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.3 m across a rough floor without acceleration, if the effective coefficient of friction was 0.50?

Short answer

The value of work done is 2321.62 J.

Step-by-step solution

Draw the free body diagram of the crate

In this problem, the crate is not accelerating vertically. The forces acting in a vertical direction are the normal force and the weight of the crate.

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

Given data:

The mass of the crate is\(m = 46\;{\rm{kg}}\).

The height is\(h = 10.3\;{\rm{m}}\).

The coefficient of friction is\({\mu _k} = 0.5\).

The free body diagram of the crate is as follows:

The relation of force in the y-direction is given as:

\(\begin{array}\Sigma {F_{\rm{y}}} &= 0\\N - W &= 0\\N &= W &= mg\end{array}\)

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

\(\begin{array}\Sigma {F_{\rm{x}}} &= 0\\F - {F_{\rm{f}}} &= 0\\F &= {F_{\rm{f}}}\\F &= {\mu _k}mg\end{array}\)

Here, F is the pushing force, g is the gravitational acceleration, and \({F_{\rm{f}}}\) is the frictional force.

Determine the work done by the pushing force

The relation of work done is given by:

\(W = F \times h\)

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

\[\begin{array}W &= {\mu _{\rm{k}}}mg \times h\\W &= \left( {0.50} \right)\left( {46\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {10.3\;{\rm{m}}} \right)\\W &= 2321.62\;{\rm{J}}\end{array}\]

Thus, \[W = 2321.62\;{\rm{J}}\] is the required work done.