A cement block with a mass of 5.0 kg is at rest on a horizontal floor.

A force of 21 N [E] starts the block moving. Find the coefficient of static friction for the block and floor.
Fs=21N, m=5.0kg, g=9.8

1 answer

To find the coefficient of static friction, we first need to calculate the maximum static friction force that can act on the block before it starts moving. This can be done using the formula:

Fs(max) = μ_s * N

where Fs(max) is the maximum static friction force, μ_s is the coefficient of static friction, and N is the normal force acting on the block. The normal force N is equal to the force of gravity acting on the block, which can be calculated as:

N = m * g

N = 5.0 kg * 9.8 m/s^2
N = 49 N

Now we can calculate the maximum static friction force:

Fs(max) = μ_s * 49 N

Given that the applied force is 21 N [E], which is less than the maximum static friction force, we can set up the following equation:

21 N = μ_s * 49 N

Now we solve for the coefficient of static friction:

μ_s = 21 N / 49 N
μ_s = 0.429

Therefore, the coefficient of static friction between the block and the floor is 0.429.