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A block of mass m is placed at relative equilibrium on an inclined plane which itself is placed on a lift moving upwards with constant velocity straight V subscript 0. If the friction coefficient between block and plane is    mu, find the instantaneous power supplied by friction to the block?? 

a) straight mumgstraight V subscript 0sin (2theta)/2 , b)  straight V subscript 0mg sin^2straight theta, c) mgstraight V subscript 0sin(2straight theta)/2 , d) straight mustraight V subscript 0mgcos^2theta

Asked by patra04011965 26th September 2018, 10:15 PM
Answered by Expert
Answer:
Figure shows the forces acting on the the block placed on an inclined plane.
Figure also shows the velocity vector V0 due to lift movement along with the resolved components.
 
Power due to force = Force×velocity = F·V , where F and V are vectors and power is expressed in terms of dot product or scalar product.
 
Here force is friction force F = μ×N = μ×m×g×cosθ
velocity component in the direction of friction force = V0×sinθ
hence power = μ×m×g×cosθ × V0×sinθ = (1/2)×μ×V0×m×g×sin(2θ)
Answered by Expert 27th September 2018, 11:23 AM
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