Question

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.​

Answer 1

Answer:

Your answer

Explanation:

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

Explanation:

Given :

mass of coin (m) = 20g = 0.02 kg

speed (u) = 25 cm/s = 0.25 m/s

time(t) = 5s

Solution:

As the object stops after 5 second,

the acceleration is,

v = u + at

(v = 0 since, the coin stops and final velocity becomes zero)

0 = 0.25 + 5a

-0.25 = 5a

-0.25/5 = a

a = -0.05 m/s²

Now,

force of friction (F),

F = ma

F = 0.02× -0.05

F = -0.001 N