A coin of mass 20 g is pushed on a table. The coin starts moving at a speed of 25 cm/s and comes to rest in 5 s. Find the average frictional force exerted by the table on the coin.

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Solution

Step 1: Given data
Mass of coin, $m = 20 g = 20 / 1000 = 0.02 k g$
Initial velocity, $u = 25 c m / s = 0.25 m / s$
Final velocity, $v = 0$
Time, $t = 5 s$
Step 2: Formula used
$v = u - a t$ ………………….(x)
$F = m a$ ……………………..(y)
Step 3: Calculation of average frictional force (F).
Substituting the given data in equation (x), we get.
$0 = 0.25 - a (5)$
$a = - 0.05 m / s^{2}$ ( negative sign indicates that there will be retardation in the body due to frictional force).
Substituting the given value mass and magnitude of acceleration obtained above in the equation (y), we get.
$F = 0.02 \times 0.05 = 0.001 N$
Hence, the average frictional force exerted by the coin on the table is equal to $0.001 N$ .

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