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Question

Physics Question on laws of motion

In the figure a block of weight 60N60\,N is placed on a rough surface. The coefficient of friction between the block and the surface is 0.50.5. The weight WW such that the block does not slip on the surface is

A

6060 N

B

602\frac{60}{\sqrt{2}} N

C

3030 N

D

302\frac{30}{\sqrt{2}} N

Answer

3030 N

Explanation

Solution

F=μR=0.5×60N=T1=T2cos45=T2sin45=W.F =\mu R = 0.5 \times 60 N = T_1 = T_2\, cos \, 45^\circ = T_2 \,sin\, 45^\circ = W.