Question

A force of 48 newtons is required to start a 5.0 kg box moving across a horizontal concrete floor. What is the coefficient of static friction between the box and the floor? What is the coefficient of kinetic friction when the acceleration is 0.70 m/s/s?

Answer 1

1) The coefficient of static friction is 0.980

2) The coefficient of kinetic friction is 0.908

Explanation:

1)

In the first situation, the box is still at rest. There are two forces acting on the box:

- The force of push, F, forward

- The force of static friction, $F_f$

Since the box is in equilibrium we have

$F_f = F$ (1)

The value of frictional force changes from zero to a maximum value which is given by:

$F_f = \mu_s mg$ (2)

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

So, if the force F needed to put the box in motion is 48 N, it means that this is also the maximum value of the force of friction. So, we can combine eq.(1) and (2) to find the coefficient of static friction:

$\mu_s = \frac{F}{mg}$

where:

F = 48 N

m = 5.0 kg

$g=9.8 m/s^2$

Substituting,

$\mu_s = \frac{48}{(5.0)(9.8)}=0.980$

2)

In this second situation, the object is already in motion, so the equation of motion is:

$F-F_f = ma$

where

F = 48 N is the force applied forward

$a = 0.70 m/s^2$ is the acceleration of the box

$F_f = \mu_k mg$ is the force of kinetic friction, where

$\mu_k$ is the coefficient of kinetic friction

We can therefore rearrange the equation to find the coefficient:

$F-\mu_k mg = ma\\\mu_k mg = F-ma\\\mu_k = \frac{F-ma}{mg}=\frac{48-(5.0)(0.70)}{(5.0)(9.8)}=0.908$

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