Question

A 20kg block is initially at rest. A 70N force is required to set the block in motion. After the motion, a force of 60N is applied to keep the block moving with constant speed. The coefficient of static friction is:

A.0.6
B.0.52
C.0.44
D.0.35

Answer 1

Recall that the coefficient of static friction is associated with that force which is applied to just set the block in motion. This frictional coefficient is defined as the ratio of the force of static friction and the normal force acting on the block. Using this relation, plug in the appropriate values accordingly and you should get the correct solution. Remember that static friction is only responsible to make the block move and has nothing to do with maintaining the motion thus produced.

Formula Used:
Coefficient of static friction $\mu_s = \dfrac{F_s}{F_N}$

Complete answer:
We know that a frictional force opposes the state of rest or motion of a body. The extent of this opposition is determined by the coefficient of friction. If a body is at rest then the resistance to motion upon application of a force is given by the coefficient of static friction, whereas if a body is set in motion, then the frictional force that the body has to overcome to maintain its state of motion is determined by the coefficient of kinetic friction.
Usually, the coefficient of static friction is larger than that of kinetic friction, since more force is required to change the state of rest of the body to maintain an already existing state of motion of the body.
Now, the coefficient of static friction determines the amount of force that is needed to apply to overcome the normal force acting on the body and commence motion.
It is given as: $\mu_s = \dfrac{F_s}{F_N}$, where $F_s$ is the force of static friction, $F_N$ is the normal force.
In the context of our question, in order to determine the static friction, we thus consider only the force that was applied to just set the block in motion.
$F_{s} = 70\;N$
The normal reaction acting on the block is equivalent to the gravitational force acting on the block but in the opposite direction.
$F_{N} = mg = 20 \times 9.8 = 196\;N$
Therefore, the coefficient of static friction is given as:
$\mu_s = \dfrac{70}{196} = 0.35$

Therefore, the correct option would be D. 0.35N.

Note:
Remember that to find the static friction we consider only the amount of force that was applied to set the body in motion. However, if we were asked to find the coefficient of kinetic friction, then we could exclusively use the amount of force that was applied to maintain the body’s state of motion, i.e.,
$\mu_k = \dfrac{F_{k}}{F_{N}} = \dfrac{60}{196} = 0.31\;N$.
Therefore, it is important to understand which coefficient of friction is associated with what kind of motion.