Answer :

Mass of the block, m = 250 g = 0.25 kg

Initial speed of the block, u=40cm/s =0.4 m/s


Initial kinetic energy of the block at this speed = mu2/2


= 0.25 × 0.4 × 0.4 × 0.5


= 0.02 J


Final speed of the block, v = 0 m/s


Final kinetic energy of the block = mv2/2


= 0.25 × 0 × 0 × 0.5


= 0 J


Let the distance the block moves before coming to rest = s m


Coefficient of friction, μ = 0.1


Acceleration due to gravity, g=9.8 m/s2


Frictional force, F = μmg


= 0.1 × 0.25 × 9.8


= 0.245 N


Work done by the frictional force = change in kinetic energy of the block


W = 0-0.02


W = -0.02 J


Also, this work done by frictional force on the block = Fscosθ


(where θ=angle between direction of frictional force F and displacement s)


-0.02 = W = 0.245 × s × cos180°


-0.02 = -0.245s


s = 0.082 m


s = 8.2 cm


Distance before the block stops due to the frictional force = 0.082 m or 8.2 cm


And work done by the frictional force on the block = -0.02 J


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