Question

Two masses ${{m}_{1}}=5kg$ and ${{m}_{2}}=10kg$ connected by an inextensible string over a frictionless pulley are moving as shown in the figure. The coefficient of friction of the horizontal surface is 0.15. The minimum weight m that should be put on top of ${{m}_{2}}$ to stop the motion is:

A. 43.3kg
B. 10.3kg
C. 18.3kg
D. 23.3kg

Answer 1

Hint: You could make free body diagrams for both the blocks including m. Now, find the condition at which the system shows the probability of no motion. Equating the forces on either side will help to do that. Then you could substitute the given values and hence get the required answer.
Formula used:
Frictional force,
${{F}_{f}}=\mu N$

Complete answer:
In the question, we are given a spring mass system. We asked to find the minimum weight m that has to be placed on ${{m}_{2}}$ so as to stop the motion in the given system. For that, let us make the free body diagram of the two masses.
For ${{m}_{1}}$,

$T={{m}_{1}}g$ ……………………………………… (1)
Now for$m+{{m}_{2}}$,

$N=\left( m+{{m}_{2}} \right)g$
${{F}_{f}}=\mu N$
$\Rightarrow {{F}_{f}}=\mu \left( m+{{m}_{2}} \right)g$
When there is no motion in the system we could say that the frictional force balances the Tension in the string, that is,
${{F}_{f}}\ge T$
$\Rightarrow T\le \mu \left( m+{{m}_{2}} \right)g$
From (1),
$\mu \left( m+{{m}_{2}} \right)g\ge {{m}_{1}}g$
Substituting the given values, we will get,
$0.15\left( m+10 \right)\ge 5$
$\Rightarrow 0.15m+1.5\ge 5$
$\therefore m\ge 23.3kg$
Therefore, we found that if the mass m is having value that is equal to greater than 23.3kg, then the system will be at rest. Hence, the minimum mass should be 23.3kg

So,option D is found to be the correct answer.

Note:
You may be wondering why we have used the inequality sign instead of the equality sign. This is because the frictional force is known to resist the relative motion (here, between the blocks and the surface). In such a condition, the greater the frictional force is the easier it will be to maintain the system at rest.