A 20 kg sled rests on a horizontal surface. if it takes 75 N to start the sled in motion and 60 N to keep it moving at a constant speed, what are the coefficients of static and kinetic friction respectively?
explain please
F = mu m g
75 = mu (20)(9.8)
solve for static mu
60 = mu m g
60 = mu (20)(9.8)
solve for kinetic mu
thanks!
To find the coefficients of static and kinetic friction, we need to use the equations that govern the forces acting on the sled. The first equation relates the force required to move an object from rest, known as the force of static friction (F_static), to the normal force (N) and the coefficient of static friction (μ_static). The equation is:
F_static = μ_static * N
In this case, the force required to start the sled in motion is 75 N.
The second equation relates the force required to keep the object moving at a constant speed, known as the force of kinetic friction (F_kinetic), to the normal force (N) and the coefficient of kinetic friction (μ_kinetic). The equation is:
F_kinetic = μ_kinetic * N
In this case, the force required to keep the sled moving at a constant speed is 60 N.
To find the coefficients of static and kinetic friction, we need to find the normal force (N) acting on the sled. The normal force is equal to the weight of the sled, which is the force due to gravity acting on an object with mass (m) given by:
N = m * g
In this case, the mass of the sled is 20 kg. The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Now, let's calculate the normal force:
N = 20 kg * 9.8 m/s^2
N = 196 N
Using the value of the normal force, we can now calculate the coefficients of static and kinetic friction.
To find the coefficient of static friction:
F_static = μ_static * N
75 N = μ_static * 196 N
μ_static = 75 N / 196 N
μ_static ≈ 0.383
To find the coefficient of kinetic friction:
F_kinetic = μ_kinetic * N
60 N = μ_kinetic * 196 N
μ_kinetic = 60 N / 196 N
μ_kinetic ≈ 0.306
Therefore, the coefficient of static friction is approximately 0.383, and the coefficient of kinetic friction is approximately 0.306.