Question

An object of mass 2 kg moving with a constant velocity of 10 m s^–1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Answer 1

Given,

Mass of the object (m) = 2 kg

Initial velocity (u) = 10 ms⁻¹

Final velocity (v) = 0

Force of friction = 7 N

Additional opposing force = 3 N

The total opposing force acts in the opposite direction of the motion because both the friction force and the extra force oppose the motion.

Total opposing (net) force = 7 N + 3 N = 10 N

Now, as Force = mass × acceleration

`a = F/m`

Substituting, we get,

`a = (-10)/2 = -5` ms⁻²

Retardation is indicated by the negative sign.

The third equation of motion states that

`v^2 - u^2 = 2as`

or

`s = (v^2 - u^2)/(2a)`

Substituting, we get,

`s = ((0)^2 - (10)^2)/(2 xx (-5))`

= `(-100)/(-10)`

= 10 m

Hence, the object travels a distance of 10 m before coming to rest.