214k views
4 votes
A 52.5 kg skater moving initially at 2.40 m/s on rough horizontal ice comes to rest uniformly in 3.52 s due to friction from the ice. What is the coefficient of friction between the skater and the ice?

User Ersan
by
8.9k points

1 Answer

2 votes

Final answer:

To find the coefficient of friction between the skater and the ice, use the equation for uniform acceleration. The frictional force between the skater and the ice can then be calculated using the equation for frictional force. Finally, use the equation for the coefficient of friction to find the value.

Step-by-step explanation:

To find the coefficient of friction between the skater and the ice, we can use the equation for uniform acceleration. The equation is given by:

Vf = Vi + at

where Vf is the final velocity, Vi is the initial velocity, a is the acceleration and t is the time. Since the skater comes to rest, Vf is 0 m/s and Vi is 2.40 m/s. The time is given as 3.52 s.

Plugging in the values, we have:

0 = 2.40 + a(3.52)

Simplifying the equation, we get:

a = -2.40/3.52

Next, we can use the equation for frictional force:

Ff = u * m * g

where Ff is the frictional force, u is the coefficient of friction, m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).

The equation can be rearranged to solve for u:

u = Ff / (m * g)

We can find the frictional force by using the equation:

Ff = m * a

Substituting the values, we get:

Ff = 52.5 * (-2.40/3.52)

Finally, plugging in the values of Ff, m, and g into the equation for the coefficient of friction:

u = (-2.40/3.52) / (52.5 * 9.8)

Solving this equation will give us the coefficient of friction between the skater and the ice.

User Maxfax
by
8.4k points