Question

In Fig. $\mathbf{7}\mathbf{}\mathbf{-}\mathbf{}\mathbf{32}$, a constant force ${\overrightarrow{\mathbf{F}}}_{\mathbf{a}}$of magnitude $\mathbf{82}\mathbf{.}\mathbf{0}\mathbf{}\mathit{N}\mathbf{}$is applied to a $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{k}\mathit{g}\mathbf{}$shoe box at angle$\mathit{\varphi }\mathbf{=}\mathbf{50}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}$, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by ${\overrightarrow{\mathbf{F}}}_{\mathbf{a}}$when the box has moved through vertical distance $\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{150}\mathbf{}\mathit{m}$?

Short answer

Work done on the box by applied force is $4.41J$.

Step-by-step solution

Given data

1. The applied force is,$F_{\text{a}}=82.0\text{N}$.
2. Mass of boxis, $m=3.0\text{kg}$.
3. Vertical height, $h=0.150\text{m}$.

Understanding the concept

By using the concept ofchange in kinetic as well as potential energy, we can find the work done during each step.

Formula:

Potential energy,$\Delta PE=mg\Delta h$

Calculate the work done on the box

There is no change in kinetic energy as the velocity isthesame. So, the work done is only due to a change in potential energy that is work done by the force of gravity.And it is given by,

$\begin{array}{rcl}W& =& \Delta PE\\ & =& mg\Delta h\\ & & \end{array}$

Here$\Delta h$ is the change in height, which is $0.150\text{m}-0=0.150\text{m}$.

Substitute the values in the above expression, and we get,

$$\begin{gathered} W=\left(3\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(0.150\text{m}\right) \\ W=4.41·\left(1\text{kg}·{\text{m/s}}^{\text{2}}·\text{m}\times \frac{1\text{J}}{1\text{kg}·{\text{m}}^2{\text{/s}}^{\text{2}}}\right) \\ W=4.41\text{J} \end{gathered}$$

Thus, work done on the box by applied force is$4.41J$ .