A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact is \[\left( g=10\text{ }m/{{s}^{2}} \right)\] |
A) 1/2
B) 2/3
C) 3/4
D) 1/4
Correct Answer: A
Solution :
The free body diagram of the block is |
N is the normal reaction exerted by inclined plane on the block. |
Applying Newton's second law to the block along and normal to the incline. |
\[mg\,\sin \,\,45{}^\circ =T\,\,\cos \,45{}^\circ +\mu \,N\] ... (1) |
\[N=mg\,\cos \,\,45{}^\circ +T\,\,sin\,45{}^\circ \] ...(2) |
Solving we get |
\[\mu ={1}/{2}\;\] |
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