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Q69PE

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Found in: Page 264

College Physics (Urone)

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 $7501.42W$.

See the step by step solution

Step 1: Conservation of energy

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,

$KE=PE$

Step 2: Velocity when the player leaves the floor

(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,

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

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

Step 3: Average force player exerts on the ground

(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,

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

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

Step 4: Average power

(c)

The average power is,

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

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,

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

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