Answer:
Force of friction, F = 45.76 N
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
[tex]F=m\times \dfrac{v-u}{t}[/tex]
[tex]F=64\ kg\times \dfrac{0-2.81\ m/s}{3.93\ s}[/tex]
F = −45.76 N
So, the frictional force exerting on the skater is 45.76 N. Hence, this is the required solution.
