A car of mass m starts from rest and acquires a ve – Class 11 Physics

NCERT Exemplar · Q9

m

(a)

\dfrac{mv}{2}

eastward and is exerted by the car engine.

(b)

\dfrac{mv}{2}

eastward and is due to the friction on the tyres exerted by the road.

\dfrac{mv}{2}

eastward exerted due to the engine and overcomes the friction of the road.

(d)

\dfrac{mv}{2}

exerted by the engine.

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Concept understanding — Newton Second Law

Newton's Second Law: The Law That Connects Force and Motion

Imagine you're pushing a shopping cart. If you push gently, it moves slowly. Push harder, and it speeds up faster. Now imagine the cart is full of groceries — even with the same push, it accelerates much more slowly than an empty cart. This everyday experience is exactly what Newton's Second Law captures.

The Intuition First

Two things matter when you push something:

How hard you push — the force you apply.
How heavy the object is — its mass.

The harder you push, the more the object speeds up. The heavier the object, the less it speeds up for the same push. So acceleration depends on both force and mass — and in opposite ways.

Note

"Acceleration" here means any change in velocity — speeding up, slowing down, or changing direction. It's not just "going faster."

The Precise Statement

Newton's Second Law says:

The acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. The acceleration is in the same direction as the net force.

In one equation:

\vec{a} = \frac{\vec{F}_{\text{net}}}{m}

Or more commonly:

\vec{F}_{\text{net}} = m \vec{a}

Where:

$\vec{F}_{\text{net}}$ is the net force (the vector sum of all forces acting on the object) — measured in newtons (N)
$m$ is the mass of the object — measured in kilograms (kg)
$\vec{a}$ is the acceleration — measured in metres per second squared ( $\text{m/s}^2$ )

\vec{F}_{\text{net}} = m \vec{a}

What This Really Means

Force causes acceleration, not velocity. A constant net force produces constant acceleration — meaning the velocity keeps changing at a steady rate. If you stop pushing, the net force becomes zero, and acceleration becomes zero (the object continues at constant velocity — that's Newton's First Law).

Mass is a measure of inertia. The more mass an object has, the harder it is to change its motion. A truck needs a much larger force than a bicycle to achieve the same acceleration.

Direction matters. Force and acceleration are vectors — they point the same way. If you push north, the acceleration is north. If multiple forces act, you must add them as vectors to find the net force.

A Simple Example

A 2 kg block is pushed with a net force of 10 N to the right.

a = \frac{F_{\text{net}}}{m} = \frac{10\ \text{N}}{2\ \text{kg}} = 5\ \text{m/s}^2

The block accelerates at $5\ \text{m/s}^2$ to the right. Every second, its velocity increases by 5 m/s in that direction. …

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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 $\frac{mv}{2}$ 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: $\mathbf{F}_{\text{net}} = m\mathbf{a}$ . 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:

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

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

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

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

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 $F = ma$ 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.

Evaluate the options. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$p = (3t^2 + 2t + 1)\ \text{kg m s}^{-1}$
›Reveal solutionSolution

Force is the time-derivative of momentum.

Given $p=3t^2+2t+1$ ,
$F=\frac{dp}{dt}=6t+2.$
At $t=3\ \text{s}$ : …
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KEAM 2026Set eng-2026-04194 marksMCQ
$g$
›Reveal solutionSolution

Force equals momentum carried away per second: $n\,m\,v$ .

Each bullet carries momentum $mv=(0.01\ \text{kg})(20\ \text{m s}^{-1})=0.2\ \text{kg m s}^{-1}$ . …
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KEAM 2026Set eng-2026-04204 marksMCQ
$m$
›Reveal solutionSolution

Use Newton's second law $F=m\dfrac{dv}{dt}$ with $F=kt$ and integrate from rest.

Newton's second law gives $m\dfrac{dv}{dt}=kt$ . Separating variables, $dv=\dfrac{kt}{m}\,dt$ . Integrating from $0$ to $t$ with $v(0)=0$ :

$v=\dfrac{k}{m}\cdot\dfrac{t^2}{2}=\dfrac{kt^2}{2m}.$ …
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KEAM 2026Set pha-2026-0419F4 marksMCQ
$ms^{-2}$
›Reveal solutionSolution

Resultant of perpendicular forces = $\sqrt{3^2 + 4^2} = 5$ N; $m = F/a = 5/2 = 2.5$ kg.

Two perpendicular forces combine as $F = \sqrt{3^2 + 4^2} = 5$ N. With acceleration $a = 2$ m s⁻², mass …
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KEAM 2025Set eng-2025-04264 marksMCQ
$25\ ms^{-1}$
›Reveal solutionSolution

Force equals mass times acceleration, and acceleration is the change in velocity over time: $F = m\,\Delta v/t = 1\ \text{N}$ .

Given: $m = 200\ \text{g} = 0.2\ \text{kg}$ , $\Delta v = 25\ \text{ms}^{-1}$ , $t = 5\ \text{s}$ .

Acceleration: …
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KEAM 2025Set eng-2025-04284 marksMCQ
$10 \text{ ms}^{-1}$
›Reveal solutionSolution

$a=F/m=25\ \text{ms}^{-2}$ ; $v=u+at=10+25\times10=260\ \text{ms}^{-1}$ .

The acceleration produced by the force is

$a=\frac{F}{m}=\frac{50}{2}=25\ \text{ms}^{-2}.$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$300\,\text{ms}^{-1}$
›Reveal solutionSolution

Force $=$ rate of momentum ejection $=\dfrac{(0.04)(300)\times25}{60}=5\,\text{N}$ .

Momentum carried per bullet. Mass $m=40\,\text{g}=0.04\,\text{kg}$ , speed $v=300\,\text{m/s}$ :
$p = mv = 0.04\times300 = 12\\,\text{kg·m/s}.$
Rate of firing. $25$ bullets per minute $= \dfrac{25}{60}\,\text{s}^{-1}$ . …
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KEAM 2025Set eng-2025-04294 marksMCQ
$\vec{F} = \hat{i} + 2\hat{j} - 2\hat{k}$
›Reveal solutionSolution

$|\vec{F}|=3\,\text{N}$ , $a=2\,\text{m/s}^2$ , so $m=F/a=1.5\,\text{kg}$ .

Magnitude of the force. For $\vec{F}=\hat{i}+2\hat{j}-2\hat{k}$ ,

$|\vec{F}| = \sqrt{1^2+2^2+(-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3\\,\text{N}.$ …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$F_1, F_2$
›Reveal solutionSolution

In equilibrium the three forces sum to zero, so $\vec F_1+\vec F_2=-\vec F_3$ . Take away $F_3$ and the leftover resultant has magnitude $F_3$ ; acceleration $=F_3/m$ .

Derivation: Equilibrium requires
$\vec F_1+\vec F_2+\vec F_3 = 0 \;\Rightarrow\; \vec F_1+\vec F_2 = -\vec F_3.$
When $F_3$ is removed, the net force is

$\vec F_{net} = \vec F_1+\vec F_2 = -\vec F_3,$ …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$Q. If the forces acting on two bodies of masses 2 kg and 3 kg are same , then the ratio of their respective accelerations is (A) 1 : 1 (B) 1 : 2 (C) 2 : 3 (D) 3 : 2 (E) 4 : 9$
›Reveal solutionSolution

From $F=ma$ with equal $F$ , $a \propto 1/m$ . So $a_1:a_2 = \dfrac{1}{2}:\dfrac{1}{3} = 3:2$ . …

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KEAM 2025Set pha-2025-0429F4 marksMCQ
$x = 3t + 7t^2$
›Reveal solutionSolution

Differentiating $x = 3t + 7t^2$ twice gives $a = 14\,ms^{-2}$ ; with $m = 0.2$ kg, $F = ma = 2.8\,\text{N}$ .

The displacement is $x = 3t + 7t^2$ . Velocity and acceleration are

$v = \frac{dx}{dt} = 3 + 14t, \qquad a = \frac{dv}{dt} = 14\,ms^{-2}.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$4\,\mathrm{ms}^{-2}$
›Reveal solutionSolution

Buoyancy is fixed at $0.84$ N; to descend at $4\ \mathrm{ms^{-2}}$ the total mass must be 140 g, so add 80 g.

Step 1: Moving up with acceleration $a=4\ \mathrm{ms^{-2}}$ : $B-mg=ma\Rightarrow B=m(g+a)=0.060\times14=0.84$ N.

Step 2: To descend with the same acceleration, weight now exceeds buoyancy: $Mg-B=Ma\Rightarrow B=M(g-a)=M\times6$ . …
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Q.A car of mass $m$ starts from rest and acquires a velocity along east $\mathbf{v} = v\hat{i}\,(v > 0)$ in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is

Concept understanding — Newton Second Law

Newton's Second Law: The Law That Connects Force and Motion

The Intuition First

The Precise Statement

What This Really Means

A Simple Example

Concept: Kinematics for the value, Newton's third law for the source

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

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

Unlock everything free for 14 days