Answer to Question #92731 in Mechanics | Relativity for JJ Lee

Question

A 41.2-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocity of -2.40 m/s. Her hands are in contact with the wall for 0.535 s. Ignore friction and wind resistance. Find the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her). Note that this force has direction, which you should indicate with the sign of your answer.

Answer 1

Impulse of the skater $J=mV$ , where m - mass of the skater and V - speed of the skater.

$J=Ft$, where F - average force that the wall applies to the skater, t - time of contact with the wall.

$mV=Ft,\ F=mV/t=41.2\cdot(-2.4)/0.535=-184.82N$.

According to the Third Newton's law the force applied by the skater is - F = 184.82N

Answer: average force that the wall applies to the skater is -184.82N, the force applied by the skater is 184.82N.