JEE MAIN - Physics (2018 (Offline) - No. 21)
Two masses m1 = 5 kg and m2 = 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 m2 to stop the motion is :
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10.3 kg
18.3 kg
27.3 kg
43.3 kg
Explanation
Moving block will stop when the friction force between m2 and surface is $$ \ge $$ tension force.
So condition for stopping the moving block,
$$f \ge T$$
$$ \Rightarrow \mu N \ge T$$
$$ \Rightarrow \mu \left( {m + {m_2}} \right)g \ge {m_1}g$$
When m is minimum then,
$$\mu \left( {m + {m_2}} \right)g = {m_1}g$$
$$ \Rightarrow m = {{{m_1} - \mu {m_2}} \over \mu }$$
$$ \Rightarrow m = {{5 - 0.15 \times 10} \over {0.15}}$$ = 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.
So condition for stopping the moving block,
$$f \ge T$$
$$ \Rightarrow \mu N \ge T$$
$$ \Rightarrow \mu \left( {m + {m_2}} \right)g \ge {m_1}g$$
When m is minimum then,
$$\mu \left( {m + {m_2}} \right)g = {m_1}g$$
$$ \Rightarrow m = {{{m_1} - \mu {m_2}} \over \mu }$$
$$ \Rightarrow m = {{5 - 0.15 \times 10} \over {0.15}}$$ = 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.
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