A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75 N is required to set the block in motion, after which a horizontal force of 60 N is required to keep the block in moving with constant speed. Find the coefficient of static and kinetic friction between the block and the surface.

μs = 0.30

μk = 0.24

Explanation:

In order to calculate the kinetic friction and static friction between the block and the surface, you take into account that the kinetic friction is important when the block is moving and the static friction when the block is at rest.

You use the following formula to find the coefficient of static friction:

      (1)

F1 = 75N

μs: coefficient of static friction = ?

M: mass of the block = 25.0kg

g: gravitational acceleration = 9.8m/s^2

You solve for μs in the equation (1):

For the coefficient of kinetic friction you have:

      (2)

F2 = 60N

μk: coefficient of kinetic friction = ?

You solve for μk in the equation (2):

Then, you have:

coefficient of static friction = 0.30

coefficient of kinetic friction = 0.24

i wanna say it’s between b and d, leaning more towards b.

the bal will not change its velocity.