A coin of mass 20 g is pushed on a table. The coin starts moving at a speed of 25 cm/s and stops in 5 seconds. Find the force of friction exerted by the table on the coin.

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Solution

Distance covered,d = average speed x time
= (1/2)(0 + 0.25) x 5 = 0.625 m
Work done by the frictional force, W= loss of kinetic energy of the coin
= (1/2) x (0.020) x (0.25)² J
= 0.000625 J

If F = frictional force,
thenW = F * d
F = W/d = (0.000625)/(0.625)
= 0.001 N.

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A coin of mass 20 g is pushed on a table. The coin starts moving at a speed of 25 cm/s and stops in 5 seconds. Find the force of friction exerted by the table on the coin.

Distance covered,d = average speed x time = (1/2)(0 + 0.25) x 5 = 0.625 mWork done by the frictional force, W= loss of kinetic energy of the coin= (1/2) x (0.020) x (0.25)2 J= 0.000625 JIf F = frictional force, thenW = F * dF = W/d = (0.000625)/(0.625) = 0.001 N.

Distance covered,d = average speed x time
= (1/2)(0 + 0.25) x 5 = 0.625 m
Work done by the frictional force, W= loss of kinetic energy of the coin
= (1/2) x (0.020) x (0.25)² J
= 0.000625 J

If F = frictional force,
thenW = F * d
F = W/d = (0.000625)/(0.625)
= 0.001 N.