4. A 55 kg ice skater is at rest on a flat skating rink. A 198 N horizontal force is needed to set the skater in motion. However, after the skater is in mo-tion, a horizontal force of 175 N keeps the skater moving at a constant velocity. Find the coefficients of static and kinetic friction between the skates and the ice.
To solve this problem, we need to use the equations of motion and Newton's second law. Let's first analyze the situation when the skater is at rest:
1. The force required to set the skater in motion is the maximum static friction force, which can be represented as F_static = u_static * N, where u_static is the coefficient of static friction and N is the normal force.
2. The weight of the skater is given by W = m * g, where m is the mass of the skater and g is the acceleration due to gravity.
3. At rest, the static friction force balances the weight of the skater, so we have F_static = W.
This can be written as: u_static * N = m * g.
Now let's analyze the situation when the skater is moving at a constant velocity:
1. The force required to keep the skater moving at a constant velocity is the kinetic friction force, which can be represented as F_kinetic = u_kinetic * N, where u_kinetic is the coefficient of kinetic friction and N is the normal force.
2. The weight of the skater is still given by W = m * g.
3. The force produced by the applied force (175 N) also contributes to the acceleration, but since the skater is moving at a constant velocity, the net force is zero.
This can be written as: F_applied + F_kinetic = 0.
Now we can solve the problem using the given information:
1. The mass of the skater is 55 kg.
2. The weight of the skater is W = m * g = 55 kg * 9.8 m/s² = 539 N.
3. The force required to set the skater in motion is F_static = 198 N.
This gives us: u_static * N = 198 N.
To find the normal force N, we can solve the equation: u_static * N = m * g.
N = (m * g) / u_static
= (55 kg * 9.8 m/s²) / u_static
= 539 N / u_static
Now let's analyze the situation when the skater is moving at a constant velocity:
1. The force required to keep the skater moving at a constant velocity is F_kinetic = 175 N.
2. The force produced by the applied force (175 N) also contributes to the acceleration, but since the skater is moving at a constant velocity, the net force is zero.
This gives us: F_applied + F_kinetic = 0.
Substituting the expressions for F_applied and F_kinetic, we get:
175 N + (u_kinetic * N) = 0
u_kinetic * N = -175 N
N = -175 N / u_kinetic
Since both expressions for N are equal to each other, we can set them equal to each other and solve for u_static:
(55 kg * 9.8 m/s²) / u_static = -175 N / u_kinetic
Simplifying, we get:
u_static = (-175 N / u_kinetic) * (u_kinetic / (55 kg * 9.8 m/s²))
= -175 N / (55 kg * 9.8 m/s²)
Finally, we need to take the absolute value of u_static since coefficients of friction are always positive:
u_static = abs(-175 N / (55 kg * 9.8 m/s²))
u_static ≈ 0.318
Therefore, the coefficient of static friction between the skates and the ice is approximately 0.318.
To find the coefficient of kinetic friction, we can substitute the value of u_static into the equation u_static * N = 198 N and solve for N:
(0.318) * N = 198 N
N ≈ 198 N / 0.318
N ≈ 622 N
Now substitute the value of N into the equation u_kinetic * N = -175 N and solve for u_kinetic:
u_kinetic * 622 N = -175 N
u_kinetic ≈ -175 N / 622 N
u_kinetic ≈ -0.281
Therefore, the coefficient of kinetic friction between the skates and the ice is approximately -0.281.