Question

Q5: An ice skater moving at 12 m/s coasts
to a halt in 95m on an ice surface. What is the coefficient
of (kinetic) friction between ice and skates?​

Answer 1

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077

Answer 2

Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.

Given-

velocity of the ice skater is 12 m/ sec.

Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.

$W_{f}=\bigtriangleup KE +\bigtriangleup PE$

The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,

$W_{f}=\bigtriangleup KE +0$

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,

$W_{f} =u_{f} mgx$

Here,  $u_{f}$ is friction force, $m$ is mass, $g$ is gravity and x is the distance .

Equate the value of kinetic energy and work done of friction for further result, we get,

$u_{f} mgx=\dfrac{1}{2} \times mv^2$

$u_{f} =\dfrac{1}{2gx} \times v^2$

$u_{f} =\dfrac{1}{9.8\times 95} \times 12^2$

$u_{f} =0.077$

Hence, the coefficient of kinetic friction between the ice and skates is 0.077.

For more about the friction, follow the link below-

https://brainly.com/question/13357196