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NCERT Exemplar · Q9

Q.A car of mass mm starts from rest and acquires a velocity along east v=vi^ (v>0)\mathbf{v} = v\hat{i}\,(v > 0) in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is

(a) mv2\dfrac{mv}{2} eastward and is exerted by the car engine.
(b) mv2\dfrac{mv}{2} eastward and is due to the friction on the tyres exerted by the road.
(c) more than mv2\dfrac{mv}{2} eastward exerted due to the engine and overcomes the friction of the road.
(d) mv2\dfrac{mv}{2} exerted by the engine.
Meghalaya MboseMCQ· 1mImportance★★★★★est
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The net force on the car is found from Newton's second law using the uniform acceleration; this net force is mv2\frac{mv}{2} eastward and arises from the friction force that the road exerts on the tyres.

Newton's second law tells us that the net force on an object equals its mass times its acceleration: Fnet=ma\mathbf{F}_{\text{net}} = m\mathbf{a}. The question asks about "the force exerted on the car," which means the net external force that causes the observed motion.

A common misconception is that the engine directly pushes the car forward. In reality, the engine turns the wheels, the wheels push backward on the road, and by Newton's third law the road pushes forward on the wheels through friction. It is this friction force from the road that accelerates the car.

Let me work through the kinematics and dynamics:

  1. Find the acceleration. The car starts from rest (u=0u = 0) and reaches velocity vv eastward in time t=2t = 2 s. Under uniform acceleration,

v=u+at  ⟹  v=0+a⋅2  ⟹  a=v2 (eastward).v = u + at \implies v = 0 + a \cdot 2 \implies a = \frac{v}{2} \text{ (eastward)}.

  1. Apply Newton's second law. The net force on the car is

Fnet=ma=m⋅v2=mv2 (eastward).F_{\text{net}} = ma = m \cdot \frac{v}{2} = \frac{mv}{2} \text{ (eastward)}.

  1. Identify the physical origin of this force.

    The engine delivers torque to the drive wheels. The rotating wheels exert a backward force on the road surface. By Newton's third law, the road exerts an equal and opposite force on the tyres—a static friction force directed forward (eastward). This friction is what actually accelerates the car.

    The engine does not directly exert a force on the car; it exerts torque on the axle. The force that appears in F=maF = ma for the car's center of mass is the external contact force from the road.

Watch out

Do not confuse the internal torque from the engine with the external force on the car. The engine enables the wheels to push on the road, but it is the road's reaction (friction) that accelerates the car forward.

  1. Evaluate the options. …

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