Question 156: Laws of Motion

Question

Two masses m
1
= 5 kg and m
2
= 10 kg, connected by an inextensible
string over a frictionless pulley, are moving as shown in the figure. The
coefficient of friction of horizontal surface is 0.15. The minimum
weight m that should be put on top of m
2
to stop the motion is :
![](https://gateclass.static.cdn.examgoal.net/QnesxWEJnmYijVnQK/eCK8N08Mme9lQl9W1twbQNO9fdQ5g/cLWLtQMyXwjU7k70O7FH4K/uploadfile.jpg)

Accepted Answer

Moving block will stop when the friction force between m
2
and surface is $ \ge $ tension force.
So condition for stopping the moving block,
$f \ge T$
$ \Rightarrow \mu N \ge T$
$ \Rightarrow \mu \left( {m + {m_2}} ight)g \ge {m_1}g$
When m is minimum then,
$\mu \left( {m + {m_2}} ight)g = {m_1}g$
$ \Rightarrow m = {{{m_1} - \mu {m_2}} \over \mu }$
$ \Rightarrow m = {{5 - 0.15 	imes 10} \over {0.15}}$ = 23.33 kg
So if m $ \ge $ 23.33 kg then the motion will stop. From option the minimum possible m is 27.3 kg.

Answer

10.3 kg

Answer

18.3 kg

Answer

27.3 kg

Answer

43.3 kg

Laws of Motion

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

Choose the correct answer: