Q7E - University Physics with Modern Physics

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?

Step-by-Step Solution

Answer

The frictional force exerted on the skater is 46.58 N .

Step 1: 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  

Step 2: 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)  

Step 3: 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 .