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 on a horizontal surface is 0.15. The minimum weight m that should be put on top of $m_2$ to stop the motion is.

• A. 43.3 kg
• B. 10.3 kg
• C. 18.3 kg
• D. 27.3 kg

Answer 1

The correct option is D

27.3 kg

Step-1: Given data:

Mass, $m_1=5kg$

Mass, $m_2=10kg$

Coefficient of friction of the horizontal surface = 0.15

Step-2: Calculation:

We already know that the force of friction is equal to the tension on the string,

$\therefore T=f=\mu \times \left(m_2+m\right)\times g$

$T=\mu \times \left(m_2+\hspace{0.17em}m\right)\times g=m_1\times g$

$g$ Is canceled and we are applying other values that we get,

$\Rightarrow 0.15\times (10+m)=5$

$\Rightarrow 10+m=\frac{5}{0.15}$

$m=\frac{5}{0.15}-10$

$m=23.33kg$

Now we get the minimum force that the force $m_2$ should put on 23.33 kg.

The minimum weight m that should be put on top of $m_2$ to stop the motion is 23.33 kg.