question_answer

A block of mass \[M=5\,kg\] is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force \[F=40\,\,N\] is applied, the acceleration of the block will be \[(g=10\,\,m/{{s}^{2}})\]        [MP PMT 2004]

A)             \[5.73\,\,m/{{\sec }^{2}}\]   

B)             \[8.0\,\,m/{{\sec }^{2}}\]

C)  \[3.17\,\,m/{{\sec }^{2}}\]

D)             \[10.0\,\,m/{{\sec }^{2}}\]

Correct Answer: A

Solution :

        Kinetic friction =\[{{\mu }_{k}}R\]\[=0.2(mg-F\sin 30{}^\circ )\] \[=0.2\left( 5\times 10-40\times \frac{1}{2} \right)\]\[=0.2(50-20)=6\ N\] Acceleration of the block \[=\frac{F\cos 30{}^\circ -\text{Kinetic friction}}{\text{Mass}}\] \[=\frac{40\times \frac{\sqrt{3}}{2}-6}{5}=5.73\ m/{{s}^{2}}\]