A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$, the work done on the two moving charged particles is (a) same as the same force law is involved in the two experiments. (b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens. (c) more for the case of a positron, as the positron moves away a larger distance. (d) same as the work done by charged particle on the stationary proton.

A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Co

NCERT Exemplar · Q2

d

(a) same as the same force law is involved in the two experiments.

(b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens.

(d) same as the work done by charged particle on the stationary proton.

Andhra Pradesh BieapMCQ· 1mImportance★★★★★est

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✓ Free question

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.

Work depends on both force and displacement. Although the positron experiences the same initial force as a proton, its much smaller mass gives it far greater acceleration and hence larger displacement in time $t$ . The work done on the positron is therefore more; the answer is (C).

The heart of this problem lies in understanding that work is not force alone—it is $W = \int \vec{F} \cdot d\vec{s}$ , the accumulation of force over the actual path traveled. Two particles experiencing the same repulsive Coulomb force will do different amounts of work if their masses differ, because the lighter particle accelerates more and covers more ground.

Both the proton and the positron carry charge $+e$ and are released from rest at distance $d$ from a stationary proton. The electrostatic repulsion obeys Coulomb's law:

F(r) = k \frac{e^2}{r^2},

where $r$ is the instantaneous separation. At any given position $r$ , both particles feel identical force. But their responses—their accelerations—are wildly different.

Step-by-step reasoning

Acceleration depends on mass.

Newton's second law gives $a = F/m$ . The proton has mass $m_p \approx 1836 \, m_e$ , while the positron has mass $m_e$ . At the same distance, the positron's acceleration is roughly 1836 times larger than the proton's.
Greater acceleration means greater displacement in the same time.

Starting from rest, a particle with larger acceleration will travel farther in time $t$ . The positron races away from the stationary proton much faster than a second proton would.
Work is force integrated over displacement.

W = \int_{d}^{r(t)} F(r) \, dr.

Even though $F(r)$ is the same function for both particles, the upper limit $r(t)$ —the position reached at time $t$ —is much larger for the positron. The positron samples the repulsive force over a longer stretch of its journey.

The force weakens with distance, but not fast enough to reverse the trend.

Yes, as the positron moves farther out, $F \propto 1/r^2$ drops. But the integral of force over distance—which is work—still grows because you are adding more (albeit smaller) contributions. The positron's head start in displacement more than compensates for the weakening force.

Alternatively, invoke energy: the work done equals the kinetic energy gained, $W = \frac{1}{2}m v^2$ . For the same force profile, the lighter positron reaches a higher speed and hence higher kinetic energy in time $t$ .
Why the other options fail:
- (A) claims the work is the same because the force law is the same. But work depends on displacement, not just the form of $F(r)$ .
- (B) suggests the positron does less work because the force weakens. This confuses instantaneous force with integrated work; the larger displacement dominates.
- (D) invokes Newton's third law: the moving particle exerts an equal and opposite force on the stationary proton. However, the stationary proton does not move (it is held at rest), so zero displacement means zero work done on it. Work is not the same.

Watch out

A common mistake is to think "same force $\Rightarrow$ same work." Work is $\int F \, ds$ ; if two particles experience the same $F(r)$ but travel different distances, the work differs.

Tip

When comparing work done on particles of different mass under the same force, remember: lighter particle $\Rightarrow$ larger acceleration $\Rightarrow$ larger displacement $\Rightarrow$ more work (equivalently, more kinetic energy).

✓Final answer

The correct option is (C) : more work is done on the positron because it moves a larger distance in the same time.

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 42 on this concept.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$Q. If the work done in increasing the velocity of a body by 25% is 900 J, then the work done in decreasing the velocity of the same body by 25% is (A) 700 J (B) 500 J (C) 225 J (D) 450 J$
›Reveal solutionSolution
Kinetic energy scales as $v^2$ , so a 25% increase and a 25% decrease change KE by different amounts (not symmetric) — the work values are 900 J and 700 J respectively.

Concept and Intuition

Because $KE\propto v^2$ , percentage changes in velocity do not translate into equal-and-opposite percentage changes in KE. Increasing $v$ by 25% multiplies KE by $(1.25)^2=1.5625$ , while decreasing $v$ by 25% multiplies KE by $(0.75)^2=0.5625$ — these are not mirror images, so the two work-done values differ.

Step-by-Step Solution

Let the initial KE be $K=\tfrac12mv^2$ .

Increasing velocity by 25%: $v'=1.25v$ , so $KE' = \tfrac12m(1.25v)^2 = 1.5625K$ .

Work done to increase $= KE'-K = 0.5625K = 900$ J $\Rightarrow K = \dfrac{900}{0.5625}=1600$ J.

Decreasing velocity by 25%: $v''=0.75v$ , so $KE''=\tfrac12m(0.75v)^2=0.5625K = 0.5625(1600)=900$ J.

Work done to decrease (magnitude) $= |K - KE''| = |1600-900| = 700$ J.

Common Mistakes

Assuming the work to decrease by 25% equals the work to increase by 25% (900 J) — the two are numerically different because of the quadratic dependence on $v$ .

Sign confusion — the work done by the retarding agent is negative, but the magnitude asked for is 700 J.

✓Final answer
$The correct option is (A) — 700 J.$

ANSWER: A
AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$\cos^{-1}(0.6)$
›Reveal solutionSolution
Applying $v^2=u^2+2as$ separately over the two half-lengths with their different friction coefficients gives a final speed of 5.6 m/s.

Concept and Intuition

Since the coefficient of kinetic friction changes halfway down the incline, the acceleration is piecewise-constant — the problem must be split into two stages, using the velocity at the end of stage one as the initial velocity for stage two.

Step-by-Step Solution

$\cos\theta=0.6\Rightarrow\sin\theta=\sqrt{1-0.36}=0.8$ (3-4-5 triangle).

Upper half, length $1.4$ m, $\mu_1=0.3$ : $a_1=g\sin\theta-\mu_1g\cos\theta=10(0.8)-0.3(10)(0.6)=8-1.8=6.2\ ms^{-2}$ .

Starting from rest: $v_1^2=0+2(6.2)(1.4)=17.36\ m^2s^{-2}$ .

Lower half, length $1.4$ m, $\mu_2=0.5$ : $a_2=g\sin\theta-\mu_2g\cos\theta=8-0.5(10)(0.6)=8-3=5\ ms^{-2}$ .

$v_2^2=v_1^2+2(5)(1.4)=17.36+14=31.36$ .

$v_2=\sqrt{31.36}=5.6\ ms^{-1}$ .

Common Mistakes

Using the same friction coefficient for the whole 2.8 m instead of splitting at the midpoint.

Forgetting $\sin\theta=0.8$ when only $\cos\theta=0.6$ is given.

✓Final answer
$5.6\ ms^{-1}$

ANSWER: B
AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$20\ ms^{-1}$
›Reveal solutionSolution
The work-energy theorem, with both gravity and constant air resistance doing negative work on the way up, gives a height of 10 m.

Concept and Intuition

Going up, both gravity and air resistance oppose the motion (both act downward), so the loss in kinetic energy equals the sum of the work done against both — this is a direct energy-balance shortcut, avoiding a separate calculation of a modified "effective g."

Step-by-Step Solution

Initial KE at X: $\dfrac12mv^2=\dfrac12(5)(20)^2=1000$ J.

KE at Y: given as 400 J.

Loss in KE $=1000-400=600$ J, which equals work done against gravity plus air resistance over height $h$ : $600=(mg+F_{air})h=(5\times10+10)h=60h$ .

$h=\dfrac{600}{60}=10$ m.

Common Mistakes

Forgetting to include the air-resistance work term and using only gravity (which would give $h=8$ m instead).

Sign errors treating air resistance as aiding rather than opposing motion on the way up.

✓Final answer
$The correct option is (B) — 10 m.$

ANSWER: B
AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$2\ \mathrm{Nm^{-1}}$
›Reveal solutionSolution
Work-energy theorem: the block's initial kinetic energy is dissipated by friction and stored as spring PE by the time it stops, giving $v=0.4\ \mathrm{ms^{-1}}$ — option (C).

Concept and Intuition

An impulse gives the block an initial velocity; as it slides, friction removes energy (as heat) and the spring stores energy (as elastic PE). Since the block eventually comes to rest, ALL the initial kinetic energy must have gone into these two channels — this is a direct application of the work-energy theorem over the whole sliding distance.

Step-by-Step Solution

Let $v$ be the speed just after the impulse. Initial KE $=\tfrac12mv^2$ .

Energy lost to friction over distance $d$ : $W_f=\mu mg\,d$ .

Energy stored in the spring at maximum stretch $d$ (block stops, so this is the full compression): $W_s=\tfrac12kd^2$ .

Energy conservation (block starts with KE, ends at rest with the energy fully transferred): $\tfrac12mv^2=\mu mgd+\tfrac12kd^2$ .

Substitute numbers: $\mu mgd = 0.1\times0.18\times10\times0.06=0.0108\ \mathrm J$ ; $\tfrac12kd^2=0.5\times2\times0.06^2=0.0036\ \mathrm J$ .

Sum: $0.0108+0.0036=0.0144\ \mathrm J$ . So $\tfrac12(0.18)v^2=0.0144 \Rightarrow 0.09v^2=0.0144 \Rightarrow v^2=0.16 \Rightarrow v=0.4\ \mathrm{ms^{-1}}$ .

Common Mistakes

Forgetting the friction term and equating KE only to spring PE (or vice versa) — both channels remove energy simultaneously as the block slides.

Arithmetic slip in computing $\mu mgd$ or $\tfrac12kd^2$ (careless decimal placement with $d=0.06$ ).

✓Final answer
$0.4\ \mathrm{ms^{-1}}$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$g = 10\, \mathrm{ms^{-2}}$
›Reveal solutionSolution
The 0.36 kg mass is the lighter side of an Atwood machine and rises; the string (tension) does positive work on it equal to $T\times s=8$ J in the first second — option (C).

Concept and Intuition

In an Atwood machine, the heavier mass falls and the lighter mass rises with the same magnitude of acceleration, connected by a single tension throughout the (light, inextensible) string. The work done by the string on either mass is simply tension times the distance that mass moves (positive if the tension and displacement are in the same direction, as they are here for the rising 0.36 kg mass).

Step-by-Step Solution

Let $m_1=0.36$ kg (lighter, rises) and $m_2=0.72$ kg (heavier, falls).

Acceleration of the system: $a=\dfrac{(m_2-m_1)g}{m_1+m_2}=\dfrac{(0.72-0.36)\times10}{1.08}=\dfrac{3.6}{1.08}=\dfrac{10}{3}\ \mathrm{ms^{-2}}$ .

Tension from the lighter mass's equation of motion ( $T-m_1g=m_1a$ ): $T=m_1(g+a)=0.36\left(10+\dfrac{10}{3}\right)=0.36\times\dfrac{40}{3}=4.8\ \mathrm N$ .

(Check with the heavier mass: $T=m_2(g-a)=0.72\times\dfrac{20}{3}=4.8\ \mathrm N$ — consistent.)

Distance moved by $m_1$ in the first second (starts from rest): $s=\tfrac12at^2=\tfrac12\times\dfrac{10}{3}\times1^2=\dfrac{5}{3}\ \mathrm m$ .

Tension acts upward on $m_1$ , and $m_1$ moves upward, so the work done by the string is positive: $W=Ts=4.8\times\dfrac{5}{3}=8\ \mathrm J$ .

Common Mistakes

Using the wrong mass's tension equation (mixing up which side rises and which falls).

Forgetting the string does positive work on the rising mass (tension and displacement are both upward) — the sign matters when cross-checking with the work-energy theorem including gravity.

✓Final answer
$The correct option is (C) — 8 J.$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$g=10\ ms^{-2}$
›Reveal solutionSolution
Mechanical energy is conserved as the body rises (no air resistance mentioned): the kinetic energy lost equals the gravitational PE gained. Setting the KE to half its initial value and solving for height gives $x=6\,m$ .

Concept and Intuition

As a body is thrown straight up, kinetic energy steadily converts into gravitational potential energy while total mechanical energy stays constant. So at any height $x$ : $KE(x) = KE_0 - mgx$ . We just need the height at which this has dropped to half the initial value — this is a direct energy-bookkeeping problem, no need to find the initial velocity explicitly (though we could).

Step-by-Step Solution

Given: $m = 2\,kg$ , $KE_0 = 240\,J$ , $g = 10\,ms^{-2}$ .

At height $x$ : $KE(x) = KE_0 - mgx = 240 - (2)(10)x = 240 - 20x$ .

We want $KE(x) = \dfrac{240}{2} = 120\,J$ .

$240 - 20x = 120 \Rightarrow 20x = 120 \Rightarrow x = 6\,m$ .

Common Mistakes

Trying to first solve for $v_0$ and then use kinematics equations — unnecessary extra work; direct energy conservation is faster and less error-prone.

Forgetting to halve $240$ correctly, or mixing up "half of KE lost" with "half of KE remaining."

✓Final answer
$The correct option is (C) — 6 m.$

ANSWER: C
AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$m = 0.1\ \text{kg}$
›Reveal solutionSolution
The block’s initial gravitational potential energy converts to kinetic energy, which is then dissipated by friction over repeated trips across the rough segment, with elastic bounces at the wall. The total distance traveled on the rough surface is 40 m, plus the initial 1 m smooth segment gives 41 m, but the block also travels the smooth segment each time it passes — careful accounting shows the total horizontal distance is 59 m, so the correct option is (B).

Concept and Intuition

The block starts with gravitational potential energy $mgh$ at the top of the smooth curve. Since the curve and the first horizontal segment AB are frictionless, all that energy becomes kinetic energy by the time the block reaches point B. From B onward, the rough segment BC (length 2 m) has friction, which does negative work, slowing the block. When the block hits the vertical wall at C, the collision is perfectly elastic, so the block rebounds with the same speed (but opposite direction) — no energy is lost in the bounce. Then it travels back across the rough segment, losing more energy to friction, until it either stops or reaches the smooth segment again. Because the smooth segment AB has no friction, the block coasts across it without losing energy, then re-enters the rough segment from the left. This back-and-forth continues until all the initial energy is dissipated by friction. The key is to find the total distance traveled on the rough surface, then add the distances traveled on the smooth surface (which are traversed multiple times).

Step-by-Step Solution

Find the speed at point B (just before entering the rough patch). The block starts from rest at height $h = 4\ \text{m}$ . By conservation of mechanical energy on the smooth curve and smooth AB:

$mgh = \frac{1}{2} m v_B^2 \quad \Rightarrow \quad v_B = \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 4} = \sqrt{80} = 4\sqrt{5}\ \text{m/s}.$
So the kinetic energy at B is $\frac{1}{2} m v_B^2 = mgh = 0.1 \times 10 \times 4 = 4\ \text{J}$ .

Work done by friction per unit distance.

On the rough segment, the friction force is $f = \mu m g = 0.1 \times 0.1 \times 10 = 0.1\ \text{N}$ .

The work done by friction over a distance $d$ is $W_f = -f d = -0.1 d$ (in joules). This work removes kinetic energy from the block.

First trip from B to C (2 m).

The block travels from B to C, a distance of 2 m. Friction removes $0.1 \times 2 = 0.2\ \text{J}$ of energy.

Kinetic energy remaining at C: $4 - 0.2 = 3.8\ \text{J}$ .

Speed at C: $v_C = \sqrt{2 \times 3.8 / 0.1} = \sqrt{76} \approx 8.72\ \text{m/s}$ .

The block hits the wall and rebounds elastically — speed magnitude unchanged, direction reversed.

First return trip from C to B (2 m).

The block travels back from C to B, again 2 m on rough surface. Friction removes another $0.2\ \text{J}$ .

Energy at B (on return): $3.8 - 0.2 = 3.6\ \text{J}$ .

Since AB is smooth, the block coasts across AB (1 m) without losing energy, then reaches the left end of the rough segment (point B again) with the same 3.6 J. But note: the block has now traveled 2 m (B→C) + 2 m (C→B) = 4 m on rough surface so far.

Subsequent trips: each full round trip across the rough segment (B→C→B) is 4 m and removes 0.4 J of energy.

However, the block does not necessarily always make a full round trip — it may stop partway. Let’s track the energy after each pass:

After first B→C: energy = 3.8 J

After first C→B: energy = 3.6 J

After second B→C: energy = 3.4 J

After second C→B: energy = 3.2 J

… and so on.

Each full back-and-forth (4 m) reduces energy by 0.4 J. Starting from 4 J, the number of full 4 m cycles possible is $\lfloor 4 / 0.4 \rfloor = 10$ full cycles, which would use $10 \times 4 = 40\ \text{m}$ of rough travel and reduce energy to $4 - 10 \times 0.4 = 0\ \text{J}$ . But wait — does it exactly hit zero? Let’s check: after 10 cycles (20 half-trips of 2 m each), the energy would be exactly 0. That means the block stops exactly at point B after the 20th half-trip (i.e., after returning from C to B for the 10th time). So the total rough distance is exactly 40 m.

Now account for the smooth segment AB (1 m).

Each time the block goes from B to the left end of the smooth segment and back to B, it travels 2 m on smooth surface. But careful: the block starts at A (after coming down the curve) and goes to B (1 m smooth). Then after each return to B from the rough side, it coasts across the smooth segment to the left end and back to B. How many times does it cross the smooth segment?

First time: from A to B (1 m) — this is the initial smooth travel.

Then after each return to B from the rough side (except the last one where it stops at B), the block will cross the smooth segment to the left and come back. But does it always go all the way to the left end? Since AB is frictionless, the block will coast to the leftmost point (where the curve starts) and then reverse direction (the curve is smooth, but the block would go up the curve a bit? Actually, the problem says the block is on the horizontal surface; the curve ends at A. The block cannot go up the curve because it’s not on the curve — it’s on the horizontal. So the block will slide to the left until it reaches point A, then? There is no wall at A; the block would just continue onto the curve? But the curve is smooth and upward; if the block has any speed, it would go up the curve. However, the problem likely intends that the block stays on the horizontal surface and simply reverses direction at A because the curve is a dead end? Actually, the figure shows a curved ramp that ends at A; the block comes down from the ramp and then is on the horizontal. If it slides left on the smooth segment, it would hit the base of the curve and then go up the curve. But the curve is smooth, so it would convert kinetic energy back to potential. This complicates things. Let’s re-read: "On the horizontal surface, path AB is smooth and path BC is rough". The block is released from the curved smooth surface, so it arrives at A with speed. Then it goes across AB (smooth) to B. After bouncing at C and returning to B, it will slide left on AB. Since AB is smooth and there is no wall at A, the block will go up the curved surface (which is also smooth) to some height, then come back down. That means the block will oscillate on the smooth part, but the energy lost only happens on the rough BC. So the smooth segment AB and the curve together form a frictionless system where the block can shuttle back and forth without loss. So the block will repeatedly cross AB (1 m each way) many times. But the question asks for "total distance covered by the block on the horizontal surface". The horizontal surface includes AB and BC. So we need to count all horizontal travel.

Let’s track the motion step by step:

Start at top of curve, slide down to A (this is not horizontal, so not counted).

From A to B: 1 m horizontal (smooth).

From B to C: 2 m rough.

Elastic bounce at C, then C to B: 2 m rough.

Now at B, the block has 3.6 J. It then goes left on AB: from B to A (1 m smooth). At A, it goes up the curve (not horizontal), then comes back down to A, then from A to B (1 m smooth). So that’s 2 m smooth horizontal travel (B→A and A→B).

Then again B→C (2 m rough), C→B (2 m rough), then B→A→B (2 m smooth), etc.

So each full cycle (after the first) consists of: 4 m rough (B→C→B) + 2 m smooth (B→A→B). The first cycle is: 1 m smooth (A→B) + 4 m rough (B→C→B). Then subsequent cycles: 2 m smooth + 4 m rough.

We already determined that the block makes exactly 10 round trips on the rough segment (each round trip is B→C→B, 4 m), for a total of 40 m rough. How many smooth trips? The first smooth trip is A→B (1 m). Then after each of the first 9 rough round trips? Wait: after the first rough round trip (B→C→B), the block is at B with 3.6 J. It then does a smooth round trip (B→A→B, 2 m) and then the next rough round trip. After the 10th rough round trip, the block ends at B with 0 J, so it stops at B. So it does not do a smooth trip after the last rough trip. So the number of smooth round trips is 9 (each 2 m) plus the initial 1 m. Total smooth horizontal distance = 1 + 9×2 = 19 m.

Total horizontal distance = rough (40 m) + smooth (19 m) = 59 m.

Check the options: 59 m corresponds to option (B).

Watch out

A common mistake is to forget that the block travels on the smooth segment multiple times — many students only count the initial 1 m from A to B, missing the back-and-forth trips on AB. Another pitfall is assuming the block stops exactly at C or somewhere else; careful energy accounting shows it stops exactly at B after an integer number of half-trips.

Tip

Notice that the energy loss per 2 m rough trip is exactly 0.2 J, and the initial energy is 4 J, so the number of 2 m trips is exactly 20, giving 40 m rough. Then the smooth trips are one less than the number of rough round trips (since the block stops at B after the last return). This pattern holds whenever the energy is an integer multiple of the loss per round trip.

✓Final answer
$The correct option is (B) . ANSWER: B$
AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$S = \dfrac{t^3}{3}$
›Reveal solutionSolution
Differentiating displacement gives velocity at t=2s as 4 m/s; the

work-energy theorem then gives work done = 24 J. Answer: (D).

Concept and Intuition

Rather than computing the force $F = m\,d^2S/dt^2$ and integrating $F\,dS$

over the interval, the work-energy theorem ( $W = \Delta KE$ ) lets us

shortcut straight to the answer once we know the velocity at the start and

end of the interval — since the body starts from rest ( $S=0$ at $t=0$ gives

$v=0$ too), the work done in the first 2 seconds is just the final kinetic

energy.

Step-by-Step Solution

Given $S(t) = \dfrac{t^3}{3}$ m.

Velocity: $v(t) = \dfrac{dS}{dt} = t^2$ .

At $t=0$ : $v=0$ (starts from rest). At $t=2$ s: $v = 2^2 = 4\ \text{ms}^{-1}$ .

Work-energy theorem: $W = \Delta KE = \tfrac12 m v_f^2 - \tfrac12 m v_i^2 = \tfrac12(3)(4)^2 - 0 = 24$ J.

Common Mistakes

Trying to directly integrate a variable force $F(t) = m\,a(t) = 6mt$ against displacement without recognizing that $W=\Delta KE$ gives the same answer far more quickly.

✓Final answer
$The correct option is (D) — 24 J.$

ANSWER: D
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$t=0$
›Reveal solutionSolution
This tests the work–energy theorem applied directly from a velocity–time graph — total work equals the change in kinetic energy, regardless of the path velocity takes in between.

Concept and Intuition

The work–energy theorem states that the net work done by all forces on a body equals its change in kinetic energy, $W = \Delta KE = \tfrac12 m v_f^2 - \tfrac12 m v_i^2$ . This holds true no matter how the velocity varies in between (linear decrease, reversal of direction, etc.) — only the initial and final speeds matter. Here the body starts at $+10\ \mathrm{ms^{-1}}$ and ends at $-20\ \mathrm{ms^{-1}}$ ; even though the velocity passes through zero and reverses direction, kinetic energy depends only on $v^2$ , so the sign reversal doesn't subtract — it's the magnitude that counts in $v^2$ .

Step-by-Step Solution

From the graph: at $t=0$ , $v_i = 10\ \mathrm{ms^{-1}}$ ; at $t=5\ \mathrm{s}$ , $v_f = -20\ \mathrm{ms^{-1}}$ .

Initial KE $= \tfrac12(4)(10)^2 = 200\ \mathrm{J}$ .

Final KE $= \tfrac12(4)(-20)^2 = \tfrac12(4)(400) = 800\ \mathrm{J}$ .

Work done by all forces $= \Delta KE = 800 - 200 = 600\ \mathrm{J}$ .

Common Mistakes

Assuming negative final velocity means negative final KE — kinetic energy is always $\ge 0$ since it depends on $v^2$ .

Trying to compute work via force $\times$ displacement using the (changing) slope, which is unnecessarily complicated — the work-energy theorem bypasses that entirely.

✓Final answer
$600\ J$

ANSWER: D
AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$\dfrac{-2}{9}\mu MgL$
›Reveal solutionSolution
Integrate the (shrinking) friction force over the length the chain slides, from $x=L/3$ to $x=L$ ; the result is $-\dfrac{2}{9}\mu MgL$ . Answer: (A).

Concept and Intuition

As the chain slips, the length resting on the table (and hence the normal force and friction force) keeps decreasing, so friction is NOT constant — you must integrate. The friction force at any moment is $\mu$ times the weight of only the portion still on the table, since the hanging portion contributes no normal force on the table.

Step-by-Step Solution

Let $x$ = length of chain hanging off the table at some instant; it starts at $x_0 = L/3$ and ends at $x_1 = L$ (whole chain off).

Mass on table at that instant: $m_{table} = M\dfrac{L-x}{L}$ .

Friction force (opposing the slipping motion): $f(x) = \mu m_{table}\, g = \mu\dfrac{Mg}{L}(L-x)$ .

Work done by friction (force opposes displacement $dx$ ): $W_f = -\displaystyle\int_{L/3}^{L} \mu\dfrac{Mg}{L}(L-x)\,dx$ .

Substitute $u=L-x$ : limits $u: 2L/3 \to 0$ , integral becomes $\displaystyle\int_0^{2L/3} u\,du = \dfrac{(2L/3)^2}{2}=\dfrac{4L^2/9}{2}=\dfrac{2L^2}{9}$ .

$W_f = -\mu\dfrac{Mg}{L}\times\dfrac{2L^2}{9} = -\dfrac{2}{9}\mu MgL$ .

Common Mistakes

Treating friction as constant (using the initial normal force throughout) instead of integrating as the mass on the table decreases.

Getting the starting fraction wrong — "two-thirds on the table" means the hanging part starts at $L/3$ , not $L/3$ on the table.

✓Final answer
$-\dfrac{2}{9}\mu MgL$

ANSWER: A
AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$12\ \mathrm{m\,s^{-1}}$
›Reveal solutionSolution
This tests the work–energy theorem for a ball accelerated over a known distance; the net force equals the kinetic energy gained divided by the displacement, giving 20 N.

Concept and Intuition

The hand exerts a net force on the ball while it is still in contact, over the distance the hand travels forward. Since there is no other agent doing work on the ball during this phase (we're told the net force), the work done by that net force converts entirely into the kinetic energy the ball has when it leaves the hand. This is the work–energy theorem: $W_{net} = \Delta KE$ .

Step-by-Step Solution

The ball starts at rest in the hand and leaves with speed $v = 12\ \mathrm{m\,s^{-1}}$ .

Kinetic energy gained: $\Delta KE = \tfrac12 m v^2 = \tfrac12(0.25)(12)^2 = \tfrac12(0.25)(144) = 18\ \mathrm{J}$ .

This equals the work done by the net force over the distance the hand moves, $d = 0.9\ \mathrm{m}$ : $W = F d$ .

So $F = \dfrac{18}{0.9} = 20\ \mathrm{N}$ .

Common Mistakes

Forgetting the ball starts from rest and instead trying to use only the final speed without the work–energy relation.

Confusing the distance the ball travels overall with the distance the hand (and hence the force) acts over.

✓Final answer
$The correct option is (B) — 20 N.$

ANSWER: B
AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$6\ \mathrm{kg}$
›Reveal solutionSolution
At uniform speed the net work done is zero, so the work done against friction simply equals (friction force) × (distance) = 12 J.

Concept and Intuition

When a body moves at constant velocity, its kinetic energy does not change, so by the work–energy theorem the net work done on it is zero. This means the work done by whatever force is pushing/pulling the body forward is exactly cancelled by the (negative) work done by friction — in other words, all the driving work is "spent" fighting friction. So the magnitude of work done against friction is simply the friction force times the distance moved.

Step-by-Step Solution

Convert distance: $200\ \text{cm} = 2\ \text{m}$ .

Compute kinetic friction force: $f = \mu_k N = \mu_k mg$ (normal force equals weight on a horizontal surface) $= 0.1 \times 6\ \text{kg} \times 10\ \text{m/s}^2 = 6\ \text{N}$ .

Work done against friction $W = f \times d = 6\ \text{N} \times 2\ \text{m} = 12\ \text{J}$ .

Common Mistakes

Forgetting to convert cm to m before multiplying.

Trying to bring in kinetic energy change (there is none, since speed is uniform) or acceleration (zero here).

✓Final answer
$The correct option is (A) — 12 J.$

ANSWER: A

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Q.A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$ , the work done on the two moving charged particles is

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