Chapter 7, Problem 36QAP

To determine

(a)

A graph of force versus time.

Answer to Problem 36QAP

The graph of force versus time is as below: -

  

Explanation of Solution

Given info:

  $\begin{array}{llllllllll}t\left(s\right)\hfill & 0\hfill & 1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill & 6\hfill & 7\hfill & 8910111213\hfill \\ F\left(N\right)\hfill & -20\hfill & -20\hfill & -10\hfill & 0\hfill & 10\hfill & 15\hfill & 18\hfill & 20\hfill & 252525252525\hfill \end{array}$

  $\begin{array}{llllllllll}t\left(s\right)\hfill & 1415\hfill & 16\hfill & 17\hfill & 18\hfill & 19\hfill & 20\hfill & 21\hfill & 22\hfill & 232425\hfill \\ F\left(N\right)\hfill & 2525\hfill & 25\hfill & 25\hfill & 25\hfill & 25\hfill & 25\hfill & 20\hfill & 15\hfill & 1050\hfill \end{array}$

  

To determine

(b)

The speed of object after $25\text{s}$

Answer to Problem 36QAP

The speed of object after $25\text{s}$ is $200\text{m/s}$

Explanation of Solution

The area under a force versus time curve is equal to the change in momentum of the object.

We break the graph into different parts two obtain the area so that we can calculate the momentum of the object.

since the object starts from rest, we have initial velocity of the object as zero that gives initial momentum of the object equals to zero.

  $\begin{array}{l}\text{Area}\text{=}\frac{1}{3}\left(3\text{s}\right)(-20\text{N})+\frac{1}{2}\left(\left(22\text{s}\right)+\left(12\text{s}\right)\right)\left(25\text{N}\right)=395\frac{\text{kg-m}}{\text{s}}=\Delta p\\ \Delta p=m{v}_{\text{f}}-m{v}_{\text{i}}=m{v}_{\text{f}}-0\\ v_{\text{f}}=\frac{\Delta p}{m}=\frac{\left(395\frac{\text{kg-m}}{\text{s}}\right)}{2\text{kg}}=2\times {10}^2\text{m/s}\end{array}$

Conclusion:

The speed of object is $200\text{m/s}$.