Answer:
The velocity is 
Explanation:
Given:
Force = 500N
Distance s= 0
To find :
Its velocity at s = 0.5 m
Solution:
Using the relation,
Now integrating on both sides
![\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cfrac%7Bv%5E%7B2%7D%7D%7B2%7D%5Cright%5D_%7B0%7D%5E%7B2%7D%3D%5Cleft%5B%5Cleft%2830.77%20s-19.23%20s%5E%7B2%7D%5Cright%29%5Cright%5D_%7B0%7D%5E%7B0.5%7D)
![\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cfrac%7Bv%5E%7B2%7D%7D%7B2%7D%5Cright%5D%3D%5Cleft%5B%5Cleft%2830.77%280.5%29-19.23%280.5%29%5E%7B2%7D%5Cright%29%5Cright%5D)
![\left[\frac{v^{2}}{2}\right]=[15.385-4.807]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cfrac%7Bv%5E%7B2%7D%7D%7B2%7D%5Cright%5D%3D%5B15.385-4.807%5D)
![\left[\frac{v^{2}}{2}\right]=10.578](https://tex.z-dn.net/?f=%5Cleft%5B%5Cfrac%7Bv%5E%7B2%7D%7D%7B2%7D%5Cright%5D%3D10.578)
<span>A phase change is an example of a
Physial change</span>
<span>C) Just before he starts to fall to the ground
Remember that "gravitational potential energy" is the energy the object potentially has if it were to drop from the current height. So let's look at the options and see what's correct.
A) When he initially lands
* He's currently near the lowest height. So his gravitational potential energy is quite low. So this is the wrong choice.
B) As he runs forward to plant the pole
* He's on the ground. So has almost no height. So almost no gravitational potential energy. So this too is wrong.
C) Just before he starts to fall to the ground
* Ah. He's at his maximum height. So his potential gravitational potential energy is maximized. This is the correct choice.
D) After take-off as the pole expands and lifts the vaulter toward the bar
* He's at a low height. So once again, this is a bad choice.</span>
