Question

A 40.2-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.82 m/s. Her hands are in contact with the wall for 1.16 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

force on her yields negative velocity so by third law her force on the wall is positive.
Change in Ke = (1/2)(40.2)(-1.82)^2 = work done
so
F * distance = Change in Ke
distance = average speed *time = (-1.82/2) * 1.16
so
F = - (20.1)(1.82^2) /[ 1.82 * 0.58 ]
= - (20.1)(1.82) / 0.58

Answer 2

V = Vo + a*t = -1.82.
0 + a*1.16 = -1.82,
a = -1.57 m/s^2.

F = M*a = 40.2 * (-1.57) =