A 42 -kg skater is standing still in front of a wall. By pushing against the wal

Question

A 42-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.2 m/s. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of 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).

magnitude N direction ---Select---opposite the velocity of the skaterthe same direction as the velocity of the skaterupwardnot enough information to tell

Explanation / Answer

First you must calculate the average acceleration the skater experiences.

Kinematic equation

Vf = Vi +at

Vf = final velocity

Vi = initial velocity

a= acceleration

t = time acceleration occured over

a = (Vf - Vi)/t

a = 1.2/.8

a = 1.5m/s^2

F = ma

magnitued:

F = (42)(1.5)

F = 63 newtons

An object accelerates in the direction of the net acceleration. This means that the direction of the force the wall exerts on her is in the direction of her motion, meaning the direction of the force she exerts on the wall is opposite the velocity of the skater.

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