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A block of mass m slides down an inclined plane of inclination `theta` with uniform speed The coefficient of friction between the block and the plane is `mu`. The contact force between the block and the plane is .A. `mg sin thetasqrt(1+mu^(2))`B. `sqrt((mg sin theta)^(2)+(mumgcostheta)^(2))`C. `mg sin theta`D. `mg` |
Answer» Correct Answer - D Block slides down with constant velocity Hence net force on the block is zero `f =mg sin theta, N =mg cos theta ,R = sqrt(f^(2) +N^(2))` . |
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