Question

A 68.5 kg skater moving initially at 2.4 m/s on rough horizontal ice comes to rest uniformly in 3.52 s due to friction from the ice. What force does friction exert on the skater?

Short answer

The frictional force exerted on the skater is 46.58 N .

Step-by-step solution

Given data

The mass of the skater is

m = 68.5 kg

The initial velocity of the electron is

u = 2.4 m/s

The final velocity of the skater is

v = 0 m/s

The time of retardation is

t = 3.52 s

Equations of motion and second law of motion

The initial velocity u , final velocity v , acceleration a and the time of travel t are related as

v = u + at .......(1)

According to the second law of motion, the force on an object of mass m and acceleration a is

F = ma ........(2)

Force on the skater

Let the retardation of the electron be a. From equation (1),

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}} \\ =\frac{0-2.4\mathrm{m}/\mathrm{s}}{3.52\mathrm{s}} \\ =-0.68\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The negative sign denotes retardation.

From equation (2), the force on the skater is

$$\begin{gathered} \mathrm{F}=\mathrm{ma} \\ =68.5\mathrm{kg}\times 0.68\mathrm{m}/{\mathrm{s}}^2 \\ =46.58\mathrm{N} \end{gathered}$$

Thus, the force is 46.58 N .