question_answer

A block of mass 10 kg is placed on a rough horizontal surface having coefficient of friction \[\mu =0.5\]. If a horizontal force of 100 N is applied on it, then the acceleration of the block wilt be:\[(g=10m/{{s}^{2}})\] [AIPMT 2002]

A) \[15\text{ }m/{{s}^{2}}\]

B) \[10\text{ }m/{{s}^{2}}\]

C) \[5\text{ }m/{{s}^{2}}\]

D) \[0.5\text{ }m/{{s}^{2}}\]

Correct Answer: C

Solution :

Key Idea: Apply Newton's second law along x-axis.

Free body diagram of block is:

From Newton's second law along x-axis

\[\Sigma {{F}_{x}}=ma\]

i.e.,       \[F-f=ma\]

or         \[F-\mu mg=ma\]

or         \[a=\frac{F-\mu mg}{m}\]

Given, \[F=100\text{ }N,\,\mu =0.5,\text{ }m=10\text{ }kg,\]

\[g=10m/{{s}^{2}}\]

Substituting die values in the above relation for acceleration of block,

\[a=\frac{(100)-(0.5)\,(10)\,(10)}{(10)}=5\,m/{{s}^{2}}\]