Question

A object of mass 5 g is pushed on a table. The object starts moving at a speed of 15 m/s and stops in 5 seconds. Find the force of friction exerted by the table on the object.?

Answer 1

m= 5g = 5/1000 = 0.005 kg

u= 15 m/s

v= 0 (Since stops after some time)

t= 5s

Now, F= ma (second law of motion)

F= 0.005 x 0-15/5

F= 5/1000 x (-3)

F= -0.015 N

Here negative sign says that the force exerted is opposite the motion of the object that is frictional force. therefore the frictional force exerted is -0.015 Newton.

Answer 2

Answer:

- 0.015N

Explanation:

mass = m = 5g = 5/1000 = 0.005 Kg (given)

Initial Velocity = $u$ = 15 m/s (given)

Final Velocity = $v$ = 0 m/s (since it stops after some time)

Time = t = 5 s (given)

Acceleration = a = $v$ - $u$ / t

                           = 0 - 15/ 5

                           =  -15/5

                           = -3 m/s²

                     

To find  force of friction exerted by the table on the object,

F = ma

F = 0.005 × -3

F = - 0.015

∴ Force of friction exerted by the table on the object is -0.015N.

Hope it Helps :)