Laws Of Motion Question 124
Question: A block of mass $ M=5kg $ is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force $ F=40N $ is applied, the acceleration of the block will be $ (g=10m/s^{2}) $ [MP PMT 2004]
Options:
A) $ 5.73m/{{\sec }^{2}} $
B) $ 8.0m/{{\sec }^{2}} $
C)$ 3.17m/{{\sec }^{2}} $
D) $ 10.0m/{{\sec }^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Kinetic friction =$ {{\mu } _{k}}R $
$ =0.2(mg-F\sin 30{}^\circ ) $
$ =0.2( 5\times 10-40\times \frac{1}{2} ) $
$ =0.2(50-20)=6\ N $
Acceleration of the block $ =\frac{F\cos 30{}^\circ -\text{Kinetic friction}}{,} $
$ =\frac{40\times \frac{\sqrt{3}}{2}-6}{5}=5.73\ m/s^{2} $
