Question

A 48.9-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.72 m/s. Her hands are in contact with the wall for 0.555 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

Answer:

+151.55N

Explanation:

She was initially at rest with a velocity of 0m/s

She attain a velocity of -1.72m/s in the direction of the negative x axis

Using equation of motion

V = u + at

A= acceleration and t = time

V(-x) = 0 + a*t

-1.72/0.555= a

a =-3.1m/s^2

The resultant that propel her away from the wall = m* a = 48.9* (-3.1) = - 151.55N

Using Newton's third law of action and equal opposite reaction force;

The wall will experience a + 151.55N force from the skater in positive x direction.