A body of mass $2\ \text{kg}$ initially at rest moves under the action of an applied horizontal force of $7\ \text{N}$ on a table with coefficient of kinetic friction $= 0.1$. Compute the (a) work done by the applied force in $10\ \text{s}$, (b) work done by friction in $10\ \text{s}$, (c) work done by the net force on the body in $10\ \text{s}$, (d) change in kinetic energy of the body in $10\ \text{s}$, and interpret your results.

A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table wi

Exercises · 5.2

2\ \text{kg}

(a) work done by the applied force in

10\ \text{s}

(b) work done by friction in

10\ \text{s}

10\ \text{s}

(d) change in kinetic energy of the body in

10\ \text{s}

,
and interpret your results.

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Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't)

It tells you: How much the speed changes when you know the net work done. Or, how much net work is needed to achieve a certain speed change.

It doesn't tell you: The direction of motion, the time taken, or the path followed. Work and kinetic energy are scalars — they have no direction.

A Quick Example

A 2 kg block initially at rest is pulled by a net force of 10 N over 4 m. Find its final speed.

Solution:

Net work: $W = Fs = 10 \times 4 = 40 \text{ J}$
Initial kinetic energy: $K_i = 0$
By the theorem: $40 = \frac{1}{2}(2)v_f^2 - 0$
So: $40 = v_f^2$
Therefore: $v_f = \sqrt{40} \approx 6.32 \text{ m/s}$

Important

The Work-Energy Theorem is a scalar alternative to Newton's laws for problems involving speed changes. It often simplifies calculations because you don't need to find acceleration or time — just work and kinetic energy.

Looking up "Work Energy Theorem: definition, formula & real-world examples" is a good habit before an exam, and it is worth knowing that Work Energy Theorem is drawn directly from the Work, Energy and Power coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers. Cross-checking this explanation against the relevant NCERT Physics chapter and solving a few past-year questions will round out your preparation.

Using the Work-Energy Theorem, we find the acceleration from net force, then displacement in 10 s. Work by applied force = 882 J, by friction = –247 J, net work = 635 J, which equals the change in kinetic energy (635 J). This confirms that net work equals change in KE.

The key to this problem is the Work-Energy Theorem: the net work done on a body equals its change in kinetic energy. But to compute individual works, we first need the displacement — which requires finding the acceleration from the net force.

Let's break it down.

1. Find the net force and acceleration

The applied force is $F_{\text{app}} = 7\ \text{N}$ forward.

Kinetic friction opposes motion: $f_k = \mu_k N$ , where $N = mg$ (since the surface is horizontal).

Mass $m = 2\ \text{kg}$ , $g = 9.8\ \text{m/s}^2$ , $\mu_k = 0.1$ .

f_k = 0.1 \times 2 \times 9.8 = 1.96\ \text{N}

Net force:

F_{\text{net}} = 7 - 1.96 = 5.04\ \text{N}

Acceleration:

a = \frac{F_{\text{net}}}{m} = \frac{5.04}{2} = 2.52\ \text{m/s}^2

Tip

Always compute friction from $N = mg$ — never assume $N = mg$ if there's a vertical force component. Here it's safe.

2. Displacement in 10 seconds

Body starts from rest ( $u = 0$ ). Using $s = ut + \frac12 a t^2$ :

s = 0 + \frac12 \times 2.52 \times (10)^2 = \frac12 \times 2.52 \times 100 = 126\ \text{m}

3. Work done by each force

Work by applied force Force and displacement are in the same direction: $W_{\text{app}} = F_{\text{app}} \cdot s = 7 \times 126 = 882\ \text{J}$
Work by friction Friction opposes motion, so $\theta = 180^\circ$ : $W_{\text{fric}} = f_k \cdot s \cdot \cos 180^\circ = 1.96 \times 126 \times (-1) = -246.96\ \text{J} \approx -247\ \text{J}$
Work by net force Either sum the works: $W_{\text{net}} = 882 + (-247) = 635\ \text{J}$ Or directly: $F_{\text{net}} \times s = 5.04 \times 126 = 635.04\ \text{J} \approx 635\ \text{J}$

4. Change in kinetic energy

Final velocity after 10 s:

v = u + at = 0 + 2.52 \times 10 = 25.2\ \text{m/s}

Initial KE = 0. Final KE:

\text{KE}_{\text{final}} = \frac12 m v^2 = \frac12 \times 2 \times (25.2)^2 = 1 \times 635.04 = 635.04\ \text{J}

So change in KE:

\Delta \text{KE} = 635.04 - 0 \approx 635\ \text{J}

5. Interpretation

Notice: $W_{\text{net}} = 635\ \text{J}$ and $\Delta \text{KE} = 635\ \text{J}$ — exactly equal. This is the Work-Energy Theorem in action.

The applied force does 882 J of work, but 247 J is "lost" to friction (converted to heat), leaving 635 J to increase the body's kinetic energy.

Watch out

A common mistake: using $W = F \times t$ instead of $W = F \times s$ . Work depends on displacement, not time. Always find $s$ first via kinematics.

W_{\text{net}} = \Delta \text{KE}

✓Final answer

The work done by applied force is 882 J, by friction is -247 J, net work is 635 J, and the change in kinetic energy is 635 J — confirming the Work-Energy Theorem.

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

KCET 2025Set D-41 markMCQ

-3

›Reveal solution

Use $W = q\,\Delta V = q\,(V_{\text{final}} - V_{\text{initial}}) = q(V_A - V_B)$ , being careful to convert millicoulombs to coulombs and to subtract the potentials in the right order (B → A).

Step 1 — The concept: potential difference is work per unit charge

The potential difference between two points is defined as the work done by an external agent in carrying a unit positive charge from one to the other, slowly (without changing its kinetic energy):

V_A - V_B = \frac{W_{B \to A}}{q}

Rearranged, this gives the working formula:

\boxed{W_{B \to A} = q\,(V_A - V_B)}

Equivalently, $W$ equals the change in electrostatic potential energy, $W = \Delta U = U_A - U_B = q V_A - q V_B$ . Since electrostatic force is conservative, this depends only on the endpoints — the path taken is irrelevant.

Step 2 — Identify the initial and final points (order matters!)

The charge is carried from B to A. So:

Initial point = B, with $V_B = +5$ V
Final point = A, with $V_A = -3$ V

The potential difference we need is $V_{\text{final}} - V_{\text{initial}} = V_A - V_B$ — not the other way round. Reversing this is the single most common error here and simply flips the sign.

V_A - V_B = (-3) - (+5) = -8~\text{V}

Step 3 — Convert the charge to SI units

q = 5~\text{mC} = 5\times 10^{-3}~\text{C}

(The prefix milli = $10^{-3}$ . Forgetting this conversion gives $-40$ J — option (D), a deliberate trap. Mistaking mC for μC would give $-4\times10^{-5}$ J.)

Step 4 — Substitute

W = q\,(V_A - V_B) = \left(5\times 10^{-3}~\text{C}\right) \times \left(-8~\text{V}\right)

W = -40 \times 10^{-3}~\text{J} = -0.04~\text{J}

\boxed{W = -0.04~\text{J}}

Step 5 — Why is the sign negative? (Check the physics)

The negative sign is not an accident — it carries real meaning.

A positive charge naturally moves from a region of higher potential to lower potential (just as a mass falls from high to low gravitational potential). Here the charge goes from $V_B = +5$ V down to $V_A = -3$ V — i.e. "downhill" in potential. The electric field itself does positive work on it; it moves there spontaneously.

Therefore the external agent does negative work: rather than pushing, the agent must hold the charge back to keep it from accelerating. Negative work by the external agent is exactly what we should expect. ✓

Step 6 — Screen the options

Option	Value	Diagnosis
(A)	$-0.04$ J	✓ Correct
(B)	$-0.4$ J	Off by a factor of 10 — arithmetic slip
(C)	$-4$ J	Off by 100
(D)	$-40$ J	The mC → C conversion was forgotten ( $5 \times -8 = -40$ )

✓Final answer

-0.04

ANSWER: A

COMEDK 2025Set 2025-E1 markMCQ
$60 \mu \mathrm{C}$
›Reveal solutionSolution
The kinetic energy gained equals the work done by the electric field: $qEd$ . Substituting the given values yields $1.2 \times 10^{-1} \, \text{J}$ , so $m = 12$ . The correct option is (A).

Concept & Intuition

When a charged particle is released from rest in a uniform electric field, the field exerts a constant force $F = qE$ on it. This force does work as the particle moves, and that work is entirely converted into kinetic energy (since no other forces act). The work done by a constant force over a displacement $d$ is simply $Fd$ . So the kinetic energy after moving distance $d$ is $K = qEd$ . No need to involve mass or acceleration — the energy approach is direct.

Step-by-step solution

Identify the given quantities

Mass $m = 3 \, \text{g} = 3 \times 10^{-3} \, \text{kg}$ (not needed for energy, but given)

Charge $q = 60 \, \mu\text{C} = 60 \times 10^{-6} \, \text{C} = 6 \times 10^{-5} \, \text{C}$

Electric field $E = 10^5 \, \text{N/C}$

Distance moved $d = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}$

Work done by the electric field

The force on the particle is $F = qE$ . Since the field is uniform, the force is constant. The work done over displacement $d$ is:

$W = Fd = qEd$
Substitute the values:
$W = (6 \times 10^{-5}) \times (10^5) \times (2 \times 10^{-2})$

Simplify step by step First, $6 \times 10^{-5} \times 10^5 = 6 \times 10^{0} = 6$ . Then multiply by $2 \times 10^{-2}$ :

$W = 6 \times 2 \times 10^{-2} = 12 \times 10^{-2} \, \text{J}$

Relate to the given form The problem states the kinetic energy is $m \times 10^{-2} \, \text{J}$ . Comparing:

$m \times 10^{-2} = 12 \times 10^{-2}$
Hence $m = 12$ .

Watch out

A common mistake is to involve the mass and compute acceleration, then velocity, then kinetic energy — that works but is unnecessarily long. The energy method is simpler and avoids unit errors.

Tip

Always check units: charge in coulombs, field in N/C, distance in metres gives work in joules directly. No need to convert mass at all here.

✓Final answer
$The correct option is (A). ANSWER: A$
COMEDK 2025Set 2025-M1 markMCQ
$24 \mathrm{Nm}^{-1}$
›Reveal solutionSolution

Total fall $=2+0.5=2.5\,$ m, so gravity does $mgh=12.5\,$ J while $\tfrac12kx^2=3\,$ J is stored in the spring; the net work done in the process is $12.5-3=9.5\,$ J — option (D).

Concept. As the ball drops onto the spring and compresses it, gravity does positive work over the whole fall, and part of that energy is stored as elastic potential energy in the spring. The net work done in the process is the work delivered by gravity minus the energy locked up in the spring.

Step 1 — Total vertical drop.

The ball starts $2\,$ m above the free (upper) end of the spring and the spring is compressed by $50\,\text{cm}=0.5\,$ m, so the ball descends
$h = 2 + 0.5 = 2.5\ \text{m}.$
Step 2 — Work done by gravity.

With $m=0.5\,$ kg and $g=10\,\text{m s}^{-2}$ ,
$W_{\text{grav}} = mgh = 0.5\times 10 \times 2.5 = 12.5\ \text{J}.$
Step 3 — Elastic energy stored in the spring.

With $k=24\,\text{N m}^{-1}$ and $x=0.5\,$ m,
$U_{\text{spring}} = \tfrac12 k x^2 = \tfrac12\times 24 \times (0.5)^2 = 3\ \text{J}.$
Step 4 — Net work done in the process.
$W_{\text{net}} = W_{\text{grav}} - U_{\text{spring}} = 12.5 - 3 = 9.5\ \text{J}.$
✓Final answer
$= 9.5\ \text{J}$
COMEDK 2024Set 2024-E1 markMCQ
$7 \mathrm{~ms}^{-1}$
›Reveal solutionSolution
Using the work–energy theorem, the braking force does work equal to the loss in kinetic energy. The computed resistance is one‑quarter of the scooter’s weight, so the correct option is (B).

The key idea is that the brakes apply a constant resisting force (the total resistance to motion) over the stopping distance. The work done by that force equals the change in kinetic energy. Because the weight $W$ is given, we can express the mass as $m = W/g$ and then solve for the force in terms of $W$ .

Identify the known quantities

Initial speed: $u = 7\ \mathrm{m/s}$

Stopping distance: $s = 10\ \mathrm{m}$

Final speed: $v = 0$

Weight of scooter: $W$ (in newtons)

Acceleration due to gravity: $g = 9.8\ \mathrm{m/s^2}$ (we will keep it symbolic initially).

Relate force to acceleration using kinematics

From $v^2 = u^2 + 2as$ , we get

$0 = (7)^2 + 2a(10) \quad\Rightarrow\quad 49 + 20a = 0 \quad\Rightarrow\quad a = -\frac{49}{20} = -2.45\ \mathrm{m/s^2}.$
The negative sign means deceleration. The magnitude of the acceleration is $a = 2.45\ \mathrm{m/s^2}$ .

Apply Newton’s second law The total resisting force $F$ (assumed constant) is

$F = m a,$
where $m$ is the mass. Since weight $W = mg$ , we have $m = W/g$ . Thus
$F = \frac{W}{g} \cdot a.$

Substitute numbers Using $g = 9.8\ \mathrm{m/s^2}$ and $a = 2.45\ \mathrm{m/s^2}$ :

$F = \frac{W}{9.8} \times 2.45 = W \times \frac{2.45}{9.8} = W \times \frac{1}{4}.$
So $F = \frac{1}{4}W$ .

Tip

Notice that $2.45 = 9.8/4$ , so the ratio is exactly $1/4$ without any rounding. This neat fraction arises because $7^2 = 49$ and $2 \times 10 = 20$ , giving $a = 49/20 = 2.45$ , and $g = 9.8 = 49/5$ . The cancellation is clean.

Interpret the result The total resistance (braking force) is one‑quarter of the scooter’s weight. This matches option (B).

Watch out

A common mistake is to forget that weight is a force ( $W = mg$ ) and to treat $W$ as mass. Always write $m = W/g$ before using $F = ma$ .

✓Final answer
$The correct option is (B) . ANSWER: B$

←Previous question5.1 — The sign of work done by a force on a body is important to understand.…→Next question5.3 — Given in Fig. 5.11 are examples of some potential energy functions in …

Q.A body of mass $2\ \text{kg}$ initially at rest moves under the action of an applied horizontal force of $7\ \text{N}$ on a table with coefficient of kinetic friction $= 0.1$ . Compute the

Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

The Intuition First

The Precise Statement

Why "Net" Work Matters

A Simple Derivation (for constant force)

What It Tells You (and What It Doesn't)

A Quick Example

1. Find the net force and acceleration

2. Displacement in 10 seconds

3. Work done by each force

4. Change in kinetic energy

5. Interpretation

Concept: Work-Energy Theorem Combined with Kinematics

Step 1 — The concept: potential difference is work per unit charge

Step 2 — Identify the initial and final points (order matters!)

Step 3 — Convert the charge to SI units

Step 4 — Substitute

Step 5 — Why is the sign negative? (Check the physics)

Step 6 — Screen the options

Q.A body of mass 2 kg2\ \text{kg}2 kg initially at rest moves under the action of an applied horizontal force of 7 N7\ \text{N}7 N on a table with coefficient of kinetic friction =0.1= 0.1=0.1. Compute the

1. Find the net force and acceleration

2. Displacement in 10 seconds

3. Work done by each force

4. Change in kinetic energy

5. Interpretation

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Q.A body of mass $2\ \text{kg}$ initially at rest moves under the action of an applied horizontal force of $7\ \text{N}$ on a table with coefficient of kinetic friction $= 0.1$ . Compute the

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