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