In Fig. $\mathbf{7}\mathbf{}\mathbf{-}\mathbf{}\mathbf{33}$, a horizontal force ${\overrightarrow{\mathbf{F}}}_{\mathbf{a}}$of magnitude $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{N}\mathbf{}$is applied to a $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{k}\mathit{g}\mathbf{}$psychology book as the book slides a distance$\mathit{d}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathit{m}\mathbf{}$up a frictionless ramp at angle $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$. (a) During the displacement, what is the net work done on the book by ${\overrightarrow{\mathbf{F}}}_{\mathbf{a}}$, the gravitational force on the book, and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is its speed at the end of the displacement?

1. Force,$F_{\text{a}}=20.0\text{N}$.
2. Mass of box,$m=3.0\text{kg}$.
3. Distance traveled, $d=0.50\text{m}$.
4. The vertical angle of the ramp from the ground,$\theta =30°$ .

After resolving the applied force and weight along x and y axis, we can find its net work done on the block by the applied force and gravitational force.

From the above diagram, we can find the work done by the gravitational force and applied force.

Work done on the book by horizontal force:

As we see the diagram, we say that the component of applied force $F_a\text{cos}\theta$is along the direction of distance through which the book is moving, so work done can be calculated as,

$$\begin{gathered} W_{\text{a}}={\overrightarrow{F}}_a·\overrightarrow{d} \\ {W}_{\text{a}}=F_{a}d\text{cos}\theta \end{gathered}$$

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

$\begin{array}{rcl}{W}_{\text{a}}& =& 20\text{N×}0.50{\text{m×cos30}}^{\text{∘}}\\ & =& 8.66\left(1\text{N}\times 1\text{m}\times \frac{1\text{J}}{1\text{N}·\text{m}}\right)\\ W_{\text{a}}& =& 8.66\text{J}\end{array}$

Work done on the book by the force of gravity:

The component of weight is along the opposite direction of motion, so work done will be,

$W_{\text{g}}=Fd$

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

$W_{\text{g}}=mg\text{sin}\theta d$

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

$\begin{array}{rcl}{W}_{\text{g}}& =& 3\text{kg}\times \text{9.8}{\text{m/s}}^2\sin 30°\\ & =& -7.35\left(1\text{kg}\times 1{\text{m/s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\right)\\ W_{\text{g}}& =& -7.35\text{J}\\ & & \end{array}$

Work done by the normal force is zero. Because there is no motion ofthebook along the vertical direction,

$W_{\text{N}}=0$

Now, the total work done can be calculated as,

$W_{\text{net}}=W_{\text{a}}+W_{\text{g}}+W_{\text{N}}$

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

$$\begin{gathered} W_{\text{net}}=8.66\text{J}-7.35\text{J}+0 \\ {W}_{\text{net}}=1.31\text{J} \end{gathered}$$

Thus, the net work done on the book by the gravitational force is $1.31\text{J}$.

Now, the total work done will be equal to the change in kinetic energy as perthework-energy theorem, which can be written as,

$\Delta KE=W_{\text{net}}=\frac{1}{2}m{v}^2$

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

$\begin{array}{rcl}1.31& =& \frac{1}{2}\left(3.0\right)v^2\\ v& =& \sqrt{\frac{2.62}{3}}\\ & =& 0.935\text{m/s}\\ & & \end{array}$

Thus, the speed of the book at last displacement is$0.935\text{m/s}$.