A coin of mass $20 g$ is pushed on a table. The coin starts moving at a speed of $25 c m / s$ and stops in $5 s e c o n d s$ . Find the force of friction exerted by the table on the coin.

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Solution

Step 1: Given data
The mass of the coin $(m) = 20 g = 0.02 k g (\sin c e 1000 g = 1 k g)$
The initial speed of the coin $(u) = 25 c m / s = 0.25 m / s (\sin c e 100 c m = 1 m)$
Let the final velocity of the coin $= v m / s$
time in which coin stopped $(t) = 5 s e c o n d s$
Let the force of friction (in newton) exerted by the table $= F$
Step 02:
According to Newton's second law the rate of change of momentum is equal to the force applied.
$F = \frac{∆ p}{t} ∆ p = c h a n g e i n m o m e n t u m = m (v - u) \Rightarrow F = \frac{m (v - u)}{t} - - - e q u a t i o n (I) S u b s t i t u t i n g t h e v a l u e o f m, v, u, a n d t i n e q u a t i o n (I) w e g e t - F = \frac{0.02 (0 - 0.25)}{5} F = \frac{- 0.005}{5} F = - 0.001 N$
Hence the frictional force exerted by the table on the coin is $0.001 N$ . The negative sign shows that the direction of force is opposite to the direction of motion of the coin.

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

A coin of mass $20 g$ is pushed on a table. The coin starts moving at a speed of $25 c m / s$ and stops in $5 s e c o n d s$ . Find the force of friction exerted by the table on the coin.