Question

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

Answer 1

To answer this question, we can begin by first finding the acceleration (a) of the coin when it stops. 

Magnitude of Initial velocity of the coin (u)= 25 cm/s = 0.25 m/s
Magnitude of final velocity of the coin (v)= 0m/s (since the coin stops)
Time period under consideration (t)= 5 sec

Now, assuming uniform acceleration, using the equation of motion,

v= u + at

Therefore, a= (0-0.25)/5 m/s²= -0.05 m/s²

Now, using Newton's second law,

Net Force (F)= Mass (m) x Acceleration (a)

Note that the net force on the coin in the horizontal direction (along the table surface) is equivalent to the force of friction itself since no other horizontal force acts on the coin, and is given by:

Force of Friction= 0.02 x (-0.05) N = - 0.0010 N 

The negative sign indicates that the friction operates in a direction opposite to the direction of the coin's initial velocity. 

Answer 2

my answer will not help you very well but the first and this answer it is useful for you.( not sure)