Question

A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player moves at a speed of 3.70 m/s before coming to rest. The coefficient of friction between the player and the ground is 0.700.
a) Find the change in mechanical energy of the player (in J) from the start of the slide to when he comes to rest.
b) How far (in m) does the player slide?

Answer 1

Answer:

(a) -472.305  J

(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by$0.5mv^{2}$

Initial kinetic energy is $0.5\times 69\times 3.7^{2}=472.305 J$

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305  J=-472.305  J

(b)

From fundamental kinematic equation

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

$s=\frac {v^{2}-u^{2}}{-2a}$ but a=\mu g hence

$s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m$