Question

A 64.0 kg skater moving initially at 2.81 m/s on rough horizontal ice comes to rest uniformly in 3.93 s due to friction from the ice. What force does friction exert on the skater?

Answer 1

Force of friction, F = 45.76 N

Step-by-step explanation:

t is given that,

Mass of the skater, m = 64 kg

Initial velocity of the skater, u = 2.81 m/s

Finally it comes to rest, v = 0

Time, t = 3.93 s

We need to find the force of friction. According to seconds law of motion as :

F = m × a

`F=m* (v-u)/(t)`

`F=64\ kg* (0-2.81\ m/s)/(3.93\ s)`

F = −45.76 N

So, the frictional force exerting on the skater is 45.76 N. Hence, this is the required solution.