Question

a block of mass 0.5 kg has an initial velocity of 10 m/s and inclined plane of angle 30° ,the coefficient of friction between the block and the incline surface is 0.2 the velocity of the block after its travel a distance of 10 metre is​

Answer 1

Answer:

13 m/s (approx)

Explanation:

Doward acceleration a = gsin30°-mu×gxcis30° =10×1/2-0.2×10×√3/2 =5-√3 m/s^2

So velocity after moving 10 m downward v^2=u^2+2×a×10 =>v^2= 10^2+2×(5-√3)×10 =

v=13m/s (approx)