Question

If you use a horizontal force of 32.0 N to slide a 12.5 kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?

Answer 1

Answer:

Value of coefficient of kinetic friction is 0.26 .

Explanation:

Given:

Mass of wooden crate, m = 12.5 kg.

Horizontal force to keep the block moving with constant velocity, F = 32.0 N.

Since, the block is moving with constant velocity.

So, net force experience by it is zero.

Therefore, fore of friction is equal to applied force.

Now, force of friction , $F=\mu_kN$  (  here $\mu_k$ is coefficient of kinetic friction and N is normal force)

Therefore, $\mu_kN=\mu_k\times mg=\mu_k\times 12.5 \times 9.8=122.5\times \mu_k$

Now, both forces are equal.

$122.5\times \mu_k=32\\\mu_k=\dfrac{32}{122.5}=0.26$      

The value of coefficient of kinetic friction is 0.26 .

Hence, it is the required solution.