- #1

Jazz

- 103

- 5

## Homework Statement

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

What was his power output during the acceleration phase?

##m = 105\ kg##

##d = 0.400\ m##

##h = 0.950\ m##

## Homework Equations

##P = \frac{W}{t}##

##\Delta PE_g = \Delta KE##

## The Attempt at a Solution

The problem disregards the work done by the player when first crouches, so the work I tried to find is just when he accelerates upwards (jumps).

First we know that: ##P_{output} = \frac{W_{output}}{t} = \frac{F_{exerted}d}{t}##

The velocity with which he left the floor is the same to that just before he touches the floor when landing; and after having fallen a height ##h##:

##\Delta PE_g = \Delta KE##

##mgh = \frac{1}{2}mv^2##

##v= \sqrt{2gh} = \sqrt{(2)(9.8\ m/s^2)(0.950\ m)} = 4.32\ m/s##Assuming that the velocity was obtained after a net constant acceleration along a distance ##d##, then:

##a = \frac{v^2}{2d} = \frac{2gh}{2d} = \frac{(9.8\ m/s^2)(0.950\ m)}{0.400\ m} = 23.3\ m/s^2##Then, I try to find the average force exerted:

##F_{exerted} = F_{net} + w##

##F_{exerted} = m(a + g)##

##F_{exerted} = (105\ kg)[23.3\ m/s^2 + 9.8\ m/s^2] = 3473\ N##Then, I tried to find the time:

##t = \frac{v-v_0}{a} = \frac{4.32\ m/s}{23.3\ m/s^2} = 0.185\ s##And finally the power:

##P_{output} = \frac{F_{exerted}d}{t} = \frac{(3473\ N)(0.400\ m)}{0.185\ s} = 7.49\ kW##

But the answer provided by the textbook says that the power must be ##8.93\ kW##. After playing around with the numbers, I noticed that what makes the difference between both results is the way the time was calculated. According to the textbook, the time is:

##t= \sqrt{\frac{2d}{(a + w)}} = \sqrt{\frac{2(0.400\ m)}{(23.3\ m/s^2 + 9.8\ m/s^2)}} = 0.156\ s##

My question is: why should ##g## be added to the ##a##? It's like considering an absolute acceleration exerted by the player, but that should underestimate the time the force was exerted.

In a hypothetical situation, if I were pushing a box with an acceleration of ##1\ m/s## through a distance ##d##, and it happens that I know friction is exerting an acceleration of ##3\ m/s## in the opposite direction, the time I spent pushing the box is dependent of ##a## but not ##a + a_{friction}##. The latter will yield a shorter time than that I really spent pushing the box (and hence, a greater power output).

I'm puzzled.