A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 cm/s. If the friction coefficient between the table and the block is 0.1, how far does the block move before coming to rest?
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