A car of mass m starts from rest and acquires a velocity along east `upsilon = upsilonhati (upsilon gt 0)` in two seconds Assuming the car moves with unifrom acceleration the force exerted on the car is .

Options

A. `(mv)/(2)` eastward and is exerted by the car engine
B. `(mv)/(2)` eastward and is due to the friction on the tyres exerted by the road
C. More than `(mv)/(2)` eastward exerted due to the engine and overcomes the friction of the road
D. `(mv)/(2)` exerted by the engine

Answer

B

Explanation

Given, mass of the car = m
As car starts from rest, u = 0
Velocity acquired along east `= v vec(i)`
Duration = t = 2s
We know that `v = u + at`
`rArr v hat(i) = 0 + a xx 2`
`rArr a = (v)/(2) hat(i)`
Force , F ` = ma = (mv)/(2) hat(i)`
Hence , force acting on the car is `(mv)/(2)` towards east. As extemal force on the system is only fricition hence , the force `(mv)/(2)` is by fricition . Hence , force by engine is internal force.