Question

a coin of mass 25g is pushed on a table . The coin start moving at speed of 25cms-1 and stop at 5s . Find the force of friction exerted by the table on the coin?​

Answer 1

Given :

• Mass of the coin = 25 g = 0.025 kg
• Initial velocity (u) = 25 cm/s = 0.25 m/s
• Final velocity (v)= 0 [ stops ]
• Time taken(t) = 5 s

To find :

• Frictional force exerted by the table .

Solution :

Force is given by the formula ,

F = ΔP / t .... (1)

here ,

• F = force
• ΔP = change in momentum
• t = time taken

we know that ,

⇒ ΔP = Pբ - Pᵢ

⇒ ΔP = m v - mu

⇒ ΔP = m(v -u )

Now let's substitute the values in the above equation ,

⇒ ΔP = 0.025 ( 0 - 0.25 )

⇒ ΔP = - 0.00625 kg m/s

Now ,

⇒ F = ΔP / t

Now let's substitute the values in  equation 1 ,

⇒ F = - 0.00625/ 5

⇒ F = - 0.00123 N

The negative sign denotes that the denotes that the frictional force is opposite to the direction of motion .