Question

A 55kg ice skater is at rest on a flat skating rink. A 198N horizontal force is needed to set the skater in motion. However, after the skater is in motion, ahorizontal force of 175N keeps the skater moving at a constant velocity. Find the coefficients of static and kinetic friction between the skates and the ice.

Answer 1

initial frictional force is due to static frictional force=μsmg=horizontal force applies

μsmg=198

μs=198/55*9.8=0.367

final frictional force is due to kinematic frictional force=μkmg=horizontal force applied 175N

μkmg=175

μk=175/55*9.8

kinematic firctional coefficient=0.325

Answer 2

Let g = 10 m/s^2

Then, coefficient of st fric = 198/(10*55) = 0.36

hence,. 175/(10*55) = 0.32

Answer 3

According to Newton’s second law of motion, the net force acting on an object of mass m can be expressed as,

Here, a is acceleration.

The following is the free body diagram of the skater of mass m on ice.

$C:\Users\padmalatha\Desktop\6.gif$

In the above figure, f is the frictional force, N is normal force, and is the maximum force applied to overcome static frictional force.

When skater is set to motion there is a static friction between skater and ice and so the acceleration a of the skater is zero. To overcome the static friction, maximum force of has to be applied and then the skater will be set to motion.

Calculate the net force acting along the horizontal direction.

…… (1)

The expression for static frictional force is,

Here, is the coefficient of static friction.

Use the equation of static frictional force in equation (1).

Rearrange the equation for .

Substitute 198 N for, 55 kg for m and for g.

Thus, static frictional coefficient is.

In second case, the skater is moving with constant velocity and kinetic friction tries to keep the skater in equilibrium. Therefore the coefficient of kinetic friction between skater and ice is,

Here, is the maximum force applied.

Substitute 175 N for, 55 kg for m and for g.

Thus, kinetic frictional coefficient is.