Question

A 40-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1 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

Answer 1

The force the skater exerts is 50 N

Direction: towards wall

Step-by-step explanation:

The impulse is equal to:

`I=F_(ave) delta-t=m*delta-v=m(v_(f) -v_(i) )`

The average force is:

`F_(ave) =(m*(v_(f)-v_(i)) )/(delta-t)`

Where

vf = final speed = -1 m/s

vi = initial speed = 0

m = mass = 40 kg

Δt = time interval = 0.8 s

Replacing:

`F_(ave) =(40*(-1-0))/(0.8) =-50N` (force applied by the wall)

The force the skater exerts is

Fsk = -Fave = -(-50) = 50 N

Direction: towards wall