Question

A wooden box weighing 150 N is sliding at a constant rate over a concrete sidewalk. The force required to pull the box at a steady rate is 78 N. What is the coefficient of friction?
1.9
228
72
0.52

Answer 1

78 also then equals the force of friction (steady rate, so sum of F = 0) Ffr = mu*Fn. Normal force = weight for this box = 150. mu = 78/150 = 0.52

Answer 2

Answer:

0.52

Explanation:

The box is moving at constant velocity along the horizontal direction: this means that its acceleration is zero. According to Newton's second law:

$F_{net}=ma=0$

where $F_{net}$ is the net force acting on the object, m its mass, and a its acceleration, since the acceleration is zero, then the net force must be zero as well.

There are in total 4 forces acting on the box:

- along the vertical direction: the weight of the box (downward), given by $W=150 N$, and the normal reaction of the floor on the box (upward), $N$

- along the horizontal direction: the force that pulls the box, $F=78 N$, and the frictional force, which is given by $\mu N$, in the opposite direction

Since the net force is zero, the resultant on each direction must be zero. In the vertical direction we have:

$W-N=0\\N=W=150 N$

While in the horizontal direction we have

$F-\mu N=0\\\mu=\frac{F}{N}=\frac{78 N}{150 N}=0.52$

So, the coefficient of friction is 0.52.