A force of $50\text{ N}$ acts on an object of mass $20\text{ kg}$, directed at $30^\circ$ above the horizontal ($x$) direction, as shown in the figure. Calculate the acceleration of the object along the $x$ and $y$ directions. ![a 20 kg block pulled by a 50 N force acting at 30 degrees above the horizontal — Class 12 Physics question](/figures/tn-phy-laws-of-motion/force-50n-30deg-above-horizontal-block-20kg-components-fig-pyq-ufnfc0.svg)

A force of 50 N acts on an object of mass 20 kg, directed at 30^° above the horizontal (x) direction, as shown

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50\text{ N}

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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.

Watch out

A common mistake: thinking that a constant force means constant velocity. It doesn't — constant force means constant acceleration, so velocity keeps changing. Only when net force is zero does velocity stay constant.

Why This Law Is So Powerful

Newton's Second Law is the bridge between forces (the causes) and motion (the effects). It lets you:

Predict how an object will move if you know the forces on it
Calculate the force needed to produce a desired motion
Understand why heavier things are harder to accelerate

It applies everywhere — from a ball you throw to a rocket launching into space. The same law governs them all.

For quick revision, remember that Newton Second Law is drawn directly from the Laws of Motion coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers, which is exactly why "Newton Second Law important questions" shows up so often in Physics question banks. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.

Figure — a 20 kg block pulled by a 50 N force acting at 30 degrees above the horizontal — Class 12 Physics question — Figurea 20 kg block pulled by a 50 N force acting at 30 degrees above the horizontal — Class 12 Physics question

Step 1. Resolve $F=50$ N into components at $30°$ to the $x$ -axis: $F_x=F\cos30°=50\times0.866=43.3\text{ N}$ ; $F_y=F\sin30°=50\times0.5=25\text{ N}$ .

Step 2. Apply Newton's second law along the $x$ -axis: $a_x=\dfrac{F_x}{m}=\dfrac{43.3}{20}=2.165\text{ m s}^{-2}$ .

Step 3. Apply Newton's second law along the $y$ -axis: $a_y=\dfrac{F_y}{m}=\dfrac{25}{20}=1.25\text{ m s}^{-2}$ .

✓Final answer

a_x=2.165\text{ m s}^{-2}

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 17 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Write the mathematical equation of Newton's Second Law.$
›Reveal solutionSolution

Newton's Second Law: the net force on a body equals the rate of change of its momentum, F = dp/dt, which reduces to F = ma for constant mass.

Newton's second law states that the rate of change of momentum of a body is directly proportional to the applied net force and takes place in the direction of that force:

F = dp/dt, where p = mv is momentum.

For a body of constant mass m, dp/dt = m(dv/dt) = ma, so the familiar form is:

F = ma

where F is the net external force, m is the mass, and a is the acceleration produced.

✓Final answer
$F = dp/dt, which for constant mass gives F = ma.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the mass of a body that accelerates at a rate of 3.3 m/s^2 with a force of 120 N?$
(a) 44.36 kg
(b) 36.36 kg
(c) 54.36 kg
(d) None of these
›Reveal solutionSolution

Newton's second law, F = ma, rearranged gives m = F/a; plugging in the numbers gives about 36.36 kg.

Newton's second law: F = m a

Given F = 120 N, a = 3.3 m/s^2

m = F / a = 120 / 3.3 = 36.3636... kg approximately 36.36 kg

✓Final answer
$(b) 36.36 kg.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Match the column - select the correct definition (from Column B) for the term 'Force' (Column A):$
(a) change in linear momentum
(b) motion opposing force
(c) loss of energy
(d) rate of change of momentum
(e) ability of doing work
(f) rate of doing work
›Reveal solutionSolution

By Newton's second law, force is the rate of change of momentum.

Newton's second law of motion defines force as the time rate of change of a body's linear momentum:

F = dp/dt

For constant mass, this reduces to the familiar F = ma. This directly matches column B option (d), 'rate of change of momentum'.

✓Final answer
$Force (Column A) matches (d) rate of change of momentum (Column B).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The direction of force is always$
(a) along the velocity
(b) opposite to the velocity
(c) perpendicular to the velocity
(d) parallel to the acceleration
›Reveal solutionSolution

Newton's second law F = ma makes force parallel to acceleration. Answer (D).

Newton's second law states F = ma. Since mass m is a positive scalar, the force vector and the acceleration vector always point in the same direction; the force is parallel to the acceleration.

Force need NOT be along, opposite to, or perpendicular to velocity (for example, in circular motion the centripetal force is perpendicular to velocity, while in projectile motion gravity is neither along nor perpendicular to the velocity). But it is always parallel to the acceleration it produces.

✓Final answer
$(D) parallel to the acceleration.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Newton's second law of motion is$
(a) F = dp/dt
(b) F = mv
(c) F = mv^2
(d) F = m^2 v
›Reveal solutionSolution

Newton's second law of motion, in its most general form, is F = dp/dt (rate of change of momentum), which reduces to F = ma only when mass is constant.

Newton's second law states that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force. With the proportionality constant taken as 1 in SI units:

F = dp/dt

When mass m is constant, p = mv, so dp/dt = m(dv/dt) = ma, giving the more familiar F = ma. But the fundamental, general statement of the law -- valid even for variable-mass systems -- is F = dp/dt.

Options (b), (c), (d) are not correct forms of any physical law (F = mv, F = mv^2, F = m^2v have no standard physical meaning here).

✓Final answer
$(a) F = dp/dt.$
CBSE 2025Set ANNUAL1 markMCQ
$\vec{F} = (6\hat{i} - 8\hat{j} + 10\hat{k})\,N$
(a) $10\sqrt{2}\ \text{kg}$
(b) $2\sqrt{10}\ \text{kg}$
(c) 10 kg
(d) 20 kg
›Reveal solutionSolution

Newton's second law gives $|\vec F| = ma$ , so $m=|\vec F|/a$ ; the magnitude of the force vector is found from its components.

$\vec F = (6\hat i - 8\hat j + 10\hat k)\,N$

$|\vec F| = \sqrt{6^2+(-8)^2+10^2}=\sqrt{36+64+100}=\sqrt{200}=10\sqrt2\ N$

Given $a=1\ \text{ms}^{-2}$ , so

$m = \dfrac{|\vec F|}{a}=\dfrac{10\sqrt2}{1}=10\sqrt2\ \text{kg}$

✓Final answer
$10\sqrt2\ \text{kg}$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following quantities of a body remains constant under uniform force acting on it? (A) Velocity (B) Acceleration (C) Momentum (D) Kinetic energy$
›Reveal solutionSolution

Under a constant force, acceleration alone stays constant; velocity, momentum, KE all change.

By Newton's second law $F=ma$ , so $a=F/m$ . If the force $F$ and mass $m$ are both constant, the acceleration is constant. However, since the body keeps accelerating, its velocity $v=u+at$ keeps increasing, momentum $p=mv$ keeps increasing, and kinetic energy $\tfrac12 mv^2$ keeps increasing — none of these three remain constant.

✓Final answer
$(B) Acceleration.$
CBSE 2024Set ANNUAL1 markMCQ
$kg\,ms^{-1}$
(a) 1N
(b) 10N
(c) 1 Kgf
(d) 60m
›Reveal solutionSolution

Newton's second law states $F = \dfrac{dp}{dt}$ ; here the rate is given directly as $10\,kg\,ms^{-1}$ per second.

Newton's second law of motion states that the force acting on a body equals the rate of change of its linear momentum: $F = \dfrac{dp}{dt}$ . It is given that the momentum changes at the rate of $10\,kg\,ms^{-1}$ per second, i.e. $\dfrac{dp}{dt} = 10\,kg\,ms^{-2} = 10\,N$ (since $1\,kg\,ms^{-2} = 1\,N$ ). Therefore $F = 10\,N$ .

✓Final answer
$The correct option is (b) 10N.$
CBSE 2024Set ANNUAL1 mark
$Q. Fill in the blank: The rate of change of ______ of a body is directly proportional to the applied force.$
›Reveal solutionSolution

Newton's second law states that the rate of change of momentum is directly proportional to the applied force: F is proportional to dp/dt.

Momentum p of a body is defined as p = mv, the product of its mass and velocity. Newton's second law of motion states that the net external force acting on a body equals (is directly proportional to, and in SI units equal to) the rate of change of its momentum:

F = dp/dt = d(mv)/dt.

For constant mass this reduces to the familiar F = ma.

✓Final answer
$The blank is 'momentum' — the rate of change of momentum of a body is directly proportional to the applied force.$
CBSE 2024Set ANNUAL1 mark
$Q. Match the term with its correct dimensional formula / equation from this pool (each option is used exactly once):$
(a) [M0LT-2],
(b) [MLT-2],
(c) S = ut + (1/2)at^2,
(d) [M-1L3T-2],
(e) v = u + at,
(f) [MLT-1]. Which option correctly matches 'Force'?
›Reveal solutionSolution

Force = mass x acceleration, so its dimensions are [M][LT^-2] = [MLT^-2], matching option (b).

Force is defined by Newton's second law as F = ma. Mass has dimension [M], and acceleration has dimension [LT^-2] (rate of change of velocity, itself rate of change of displacement).

So [Force] = [M] x [LT^-2] = [MLT^-2].

Among the given options, this matches (b) [MLT^-2].

✓Final answer
$Force matches with (b) [MLT^-2].$
CBSE 2023Set ANNUAL1 markMCQ
$Q. A body of mass 2 kg travels according to the law x(t) = pt + qt^2 + rt^3, where p = 3 ms^-1, q = 4 ms^-2 and r = 5 ms^-3. The force acting on the body at t = 2 seconds is$
(a) 68N
(b) 134N
(c) 158N
(d) 136N
›Reveal solutionSolution

Differentiate x(t) twice to get a(t), then use F = ma. At t = 2 s, F = 136 N (option d).

Given x(t) = pt + qt^2 + rt^3, with p = 3 m/s, q = 4 m/s^2, r = 5 m/s^3, and mass m = 2 kg.

Step 1: Find velocity by differentiating position once.

v(t) = dx/dt = p + 2qt + 3rt^2.

Step 2: Find acceleration by differentiating velocity once (i.e., position twice).

a(t) = dv/dt = 2q + 6rt.

Step 3: Substitute t = 2 s.

a(2) = 2(4) + 6(5)(2) = 8 + 60 = 68 m/s^2.

Step 4: Apply Newton's second law.

F = ma = 2 kg x 68 m/s^2 = 136 N.

(Note the term pt contributes nothing to acceleration since it is linear in t — it only sets the initial velocity offset.)

✓Final answer
$The correct option is (d) 136 N.$
CBSE 2023Set ANNUAL1 mark
$Q. Answer in one word/sentence: Give the relation between the MKS and CGS unit of force.$
›Reveal solutionSolution

1 Newton (MKS/SI unit of force) equals 10^5 dyne (CGS unit of force).

1 Newton = 1 kg x 1 m/s^2

Convert kg to g: 1 kg = 1000 g = 10^3 g

Convert m to cm: 1 m = 100 cm = 10^2 cm

So: 1 N = 10^3 g x 10^2 cm/s^2 = 10^5 g cm/s^2

Since 1 dyne = 1 g cm/s^2 (the CGS unit of force), we get:

1 N = 10^5 dyne

✓Final answer
$1 Newton = 10^5 dyne.$

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Q.A force of 50 N50\text{ N}50 N acts on an object of mass 20 kg20\text{ kg}20 kg, directed at 30∘30^\circ30∘ above the horizontal (xxx) direction, as shown in the figure. Calculate the acceleration of the object along the xxx and yyy directions.

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Q.A force of $50\text{ N}$ acts on an object of mass $20\text{ kg}$ , directed at $30^\circ$ above the horizontal ( $x$ ) direction, as shown in the figure. Calculate the acceleration of the object along the $x$ and $y$ directions.

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