(II) What minimum horsepower must a motor have to be able to drag a 370-kg box along a level floor at a speed of 1.20 m/s if the coefficient of friction is 0.45?

In this problem, the box is dragged on the floor at a constant speed, which means that the acceleration of the box in the horizontal and vertical directions is zero.

Given data:

The mass ofthe box is\(m = 370\;{\rm{kg}}\).

The speed of the box is\(v = 1.20\;{\rm{m/s}}\).

The coefficient of kinetic friction is \({\mu _{\rm{k}}} = 0.45\).

The free body diagram of the box is as follows:

The relation of forces perpendicular to the floor is given by:

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

Here, N is the normal force, W is the weight, and g is the gravitational acceleration.

The relation of forces in a direction along the floor is given by:

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

Here,\({F_{\rm{f}}}\)is the frictional force.

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

\(F = {\mu _{\rm{k}}}mg\)

The relation to find the horsepower is given by:

\(P = Fv\)

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

\(\begin{aligned}P &= {\mu _{\rm{k}}}mg \times v\\P &= \left( {0.45} \right)\left( {370\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.20\;{\rm{m/s}}} \right)\\P &= \left( {1958.04\;{\rm{W}} \times \frac{{1\;{\rm{hp}}}}{{746\;{\rm{W}}}}} \right)\\P &= 2.62\;{\rm{hp}}\end{aligned}\)

Thus, \(P = 2.62\;{\rm{hp}}\) is the required horsepower.