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Q69PE

Expert-verifiedFound in: Page 264

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A 105-kg basketball player crouches down 0.400 m while waiting to jump. After exerting a force on the floor through this 0.400 m, his feet leave the floor and his center of gravity rises 0.950 m above its normal standing erect position. **

**(a) Using energy considerations, calculate his velocity when he leaves the floor. **

**(b) What average force did he exert on the floor? (Do not neglect the force to support his weight as well as that to accelerate him.) **

**(c) What was his power output during the acceleration phase?**

(a.) The velocity when the player leaves the floor is $4.32m/s$ .

(b) The average force exerted by the player is $3472.88N$.

(c) The average power output during the acceleration is $7501.42W$.

**The total energy of an isolated system always remains constant, only the form of energy can be changed.**

When a body is thrown upwards, the kinetic energy of the body gets converted into potential energy of the body. Mathematically,

$KE=PE$

(a)

When the player jumps, its total kinetic energy gets converted into kinetic energy i.e.,

$\frac{1}{2}m{v}_{i}^{2}=mg{h}_{f}$

Here, m is the mass of the player $\left(m=105kg\right)$, ${v}_{i}$ is the initial velocity or the velocity when the player leaves the floor, g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$ , and ${h}_{f}$ is height the center of gravity rises $\left({h}_{f}=0.950\text{m}\right)$ .

The expression for the velocity when the player leaves the floor is,

${v}_{i}=\sqrt{2g{h}_{f}}$

Putting all known values,

$\begin{array}{rcl}{v}_{i}& =& \sqrt{2\times \left(9.8\text{m}/{\text{s}}^{2}\right)\times \left(0.950\text{m}\right)}\\ & =& 4.32\text{m}/\text{s}\\ & & \end{array}$

Therefore, the required velocity when the player leaves the floor is $4.32m/s$.

(b)

The work done by the player during the jump is,

$W=mg\left({h}_{f}-{h}_{i}\right)\phantom{\rule{0ex}{0ex}}Fd=mg\left({h}_{f}-{h}_{i}\right)\phantom{\rule{0ex}{0ex}}$

Here, F is the average force exerted by the player, d is the distance $\left(d=0.400\text{m}\right)$ , m is the mass of the player$\left(m=105kg\right)$ , g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$, ${h}_{f}$ is height the center of gravity rises $\left({h}_{f}=0.950\text{m}\right)$ , and ${h}_{i}$ is the initial height or distance the player crouches down $\left({h}_{i}=-0.400\text{m}\right)$ .

The expression for the average force is,

$F=\frac{mg\left({h}_{f}-{h}_{i}\right)}{d}$

Putting all known values,

$\begin{array}{rcl}F& =& \frac{\left(105\text{kg}\right)\times \left(9.8\text{m}/{\text{s}}^{2}\right)\times \left[\left(0.950\text{m}\right)-\left(-0.400\text{m}\right)\right]}{\left(0.400\text{m}\right)}\\ & =& 3472.88\text{N}\\ & & \end{array}$

Therefore, the required average force exerted by the player is $3472.88N$ .

(c)

The average power is,

$\begin{array}{rcl}P& =& F{v}_{avg}\\ & =& F\times \left(\frac{{v}_{f}+{v}_{i}}{2}\right)\\ & & \end{array}$

Here, F is the average force exerted by the player $\left(F=3472.88\text{N}\right)$ , ${v}_{i}$ is the initial velocity of the jump $\left({v}_{i}=4.32\text{m}/\text{s}\right)$, and ${v}_{f}$ is the final velocity ( ${v}_{f}=0$ at maximum height all the kinetic energy get converted into potential energy).

Putting all known values,

$\begin{array}{rcl}P& =& \left(3472.88\text{N}\right)\times \left[\frac{\left(0\right)+\left(4.32\text{m}/\text{s}\right)}{2}\right]\\ & =& 7501.42\text{W}\end{array}$

Therefore, the required average power output during the acceleration is $7501.42\text{W}$ .

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