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.
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.
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