Question

The coefficient of kinetic friction between a 50 kg crate and the warehouse floor is 80 % of the corresponding coefficient of static friction. The crate falls off a forklift that is moving at 4 m/s and then slides along the warehouse floor for a distance of 2.5 m before coming to rest.

What is the coefficient of static friction between the crate and the floor?

Answer 1

In this answer, the value of the acceleration of gravity is taken as 9.8 m/s2

Answer:

Coefficient of static friction = 0.409

Explanation:

Kinetic friction is responsible for bringing the crate to rest.

Initial velocity, u = 4 m/s

Final velocity, v = 0 m/s

Distance, s = 2.5 m

We use the equation of motion to find the acceleration, a.

$v^2 = u^2 + 2as$

$a = \dfrac{v^2-u^2}{2s} = \dfrac{0^2-4^2}{2\times2.5} = -3.2 \text{ m/s}^2$

The negative sign denotes the crate was decelerating.

Kinetic friction = mass of crate × acceleration

Coefficient of kinetic friction = kinetic friction/weight of crate

$\mu_k=\dfrac{50\times3.2}{50\times9.8} = 0.327$

$\mu_k$ is 80% of $\mu_s$, coefficient of static friction

$0.327=0.8\mu_s$

$\mu_s = \dfrac{0.327}{0.8}=0.409$