Question

A 30 kg box sits on the floor it requires 275N of force to get it moving once it is moving it only takes 225N of force.what are the coefficient of static and kinetic friction

Answer 1

1) The coefficient of static friction is 0.935

2) The coefficient of kinetic friction is 0.765

Explanation:

1)

When a force is applied on a box sitting on the floor, the force that must be applied in order to make the box moving is equal to the maximum force of static friction between the floor and the box, which is:

$F = \mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

$g=9.8 m/s^2$

F = 275 N

Therefore, the coefficient of static friction is

$\mu_s = \frac{F}{mg}=\frac{275}{(30)(9.8)}=0.935$

2)

Once the box is in motion, the force that must be applied in order to make the box moving at constant velocity is equal to the force of kinetic friction between the floor and the box, which is:

$F = \mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

$g=9.8 m/s^2$

F = 225 N

Therefore, the coefficient of kinetic friction is

$\mu_k = \frac{F}{mg}=\frac{225}{(30)(9.8)}=0.765$

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