Question

A 100 kgblock is pulled at a constant speed of 5.0 m/sacross a horizontal floor by an applied force of 122 Ndirected${\mathbf{37}}^{\mathbf{°}}$above the horizontal. What is the rate at which the force does work on the block?

Short answer

The rate of doing work on the block is, $4.9\times {10}^2\mathrm{W}$.

Step-by-step solution

Given

The mass of block is, $\mathrm{m}=100\mathrm{kg}$.

The force applied on block is, $\mathrm{F}=122\mathrm{N}$.

The velocity of block is, $\mathrm{v}=5.0\mathrm{m}/\mathrm{s}$.

The direction of force is, $\mathrm{\theta }={37}^{°}$.

Concept

The rate at which the force does the work on the object is called as power due to the force. The work done on a particle by a constant force during its displacement is given as

$\mathrm{W}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{d}}$

Formula:

$$\begin{gathered} \mathbf{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}} \\ {\mathbf{P}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{W}}{\mathbf{∆}\mathbf{t}} \end{gathered}$$

Calculate the work done

The power due to force is the rate at which force does the work on the object.

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{avg}}& =& \frac{\mathrm{W}}{∆\mathrm{t}}\\ & =& \frac{\mathrm{Fd}\mathrm{cos\theta }}{\mathrm{t}}\\ {\mathrm{P}}_{\mathrm{avg}}& =& \mathrm{Fv}\mathrm{cos\theta }\end{array}$

(As $\frac{\mathrm{d}}{\mathrm{t}}=\mathrm{v}$),

Substitute all the value in the above equation.

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{avg}}& =& 122\mathrm{N}\times 5.0\mathrm{m}/\mathrm{s}\times \cos 37\\ & =& 4.9\times {10}^2\mathrm{W}\end{array}$

Thus, the rate at which the force does work on the block $4.9\times {10}^2\mathrm{W}$.