Question

If co efficient of friction between a fixed inclined plane and the block is 1, then the acceleration of the block shown in the following figure is (g=10m/s^2)​

Answer 1

Answer:

-1.97

Explanation:

n=1

therefore

frictional force =nmgcos37=mgcos37

now,

mgsin37-frictional force=ma

mgsin37-mgcos37=ma

m gets cancel

a=g(sin37-cos37)

a=10*-0.197

a=-1.97ms^-2

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