Work Energy Theorem And Its Applications

PhysicsWork, Energy And PowerFor JEE aspirants

The work-energy theorem states that the net work done by all forces on a body equals the change in its kinetic energy: $W_{net} = Δ K = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$ . It holds for constant or variable forces, in inertial or non-inertial frames (with pseudo-force work included), and reduces to the law of conservation of mechanical energy when only conservative forces act. This concept covers the derivation and the powerful applications for JEE Main, JEE Advanced and NEET - vertical circular motion (string, tube, rod, sphere), spring problems, friction on inclines, and motion in accelerating (non-inertial) frames.

Key Formulas - Quick Reference

Work-energy theorem: $W_{net} = Δ K = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$
Extended form: $W_{c} + W_{n c} + W_{ext} + W_{pseudo} = Δ K$
With $W_{c} = - Δ U$ : $W_{n c} + W_{ext} + W_{pseudo} = Δ (K + U) = Δ E_{mech}$
Conservation of mechanical energy (only conservative forces): $K_{i} + U_{i} = K_{f} + U_{f}$
Vertical circle (string): tension $T = \frac{m u ^{2}}{R} + 3 m g (cos θ - \frac{2}{3})$ ( $θ$ from bottom)
Critical speed at top (string): $v_{top} = g R$ ; at bottom: $v_{bottom} = 5 g R$
Tension difference: $T_{bottom} - T_{top} = 6 m g$
Rigid rod, minimum speed at bottom: $u_{m i n} = 4 g R$
String slack angle (when $2 g R < u < 5 g R$ ): $cos ϕ = \frac{u ^{2} - 2 g R}{3 g R}$
Body on outer sphere: fly-off at $cos ϕ = \frac{2}{3}$ , height $= \frac{5 R}{3}$ above the ground

1. Statement of the Work-Energy Theorem

The net work done by all forces acting on a body is equal to the change in its kinetic energy:

W_{net} = K_{f} - K_{i} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}

The theorem applies to:

Constant or variable forces
Any type of motion (linear, curvilinear, circular)
Any single particle or rigid body treated as one
Any inertial frame - and any non-inertial frame, provided the work by pseudo-forces is included

2. Derivation

2.1 One-dimensional variable force

Consider a body of mass $m$ moving along the $x$ -axis under net force $F_{net} (x)$ . From Newton's second law:

F_{net} = ma = m \frac{d v}{d t} = m v \frac{d v}{d x}

Multiplying by $d x$ and integrating from initial position $x_{i}$ (speed $v_{i}$ ) to final $x_{f}$ (speed $v_{f}$ ):

\int_{x_{i}}^{x_{f}} F_{net} d x = \int_{v_{i}}^{v_{f}} m v d v

W_{net} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = Δ K

2.2 Three-dimensional derivation

For $F_{net} = m d v / d t$ , take the dot product with $d r = v d t$ :

F_{net} \cdot d r = m v \cdot d v = m d (\frac{1}{2} v^{2})

Integrating from A to B:

W_{net} = \int_{A}^{B} F_{net} \cdot d r = \frac{1}{2} m v_{B}^{2} - \frac{1}{2} m v_{A}^{2} = Δ K

3. Extended Form and Conservation of Mechanical Energy

Classify all forces acting on the body:

W_{c} + W_{n c} + W_{ext} + W_{pseudo} = Δ K

where $W_{c}$ = work by conservative forces, $W_{n c}$ = work by non-conservative forces (friction, viscosity), $W_{ext}$ = work by applied external forces (push, pull), $W_{pseudo}$ = work by pseudo-forces in a non-inertial frame.

Since $W_{c} = - Δ U$ , this rearranges to:

W_{n c} + W_{ext} + W_{pseudo} = Δ K + Δ U = Δ E_{mech}

Special case: only conservative forces act. Then

W_{n c} = W_{ext} = W_{pseudo} = 0

, so

K_{i} + U_{i} = K_{f} + U_{f}

This is the law of conservation of mechanical energy.

4. Applications of the Work-Energy Theorem

4.1 When only one conservative force acts

Solved Example 1 (Free fall from height)

A block of mass

m

is dropped from height

h

. Find its speed on reaching the ground.

Solution:

Only gravity (conservative) acts. Using $K_{i} + U_{i} = K_{f} + U_{f}$ with reference at ground: $0 + m g h = \frac{1}{2} m v^{2} + 0 \Rightarrow v = 2 g h$

Solved Example 2 (Block compressing a spring)

A 2 kg block, attached to a spring of force constant

k

on a smooth horizontal surface, is released from rest with the spring stretched by 2 m. Find its speed when the spring is compressed by 1 m from natural length.

Solution:

Only spring force (conservative) acts. Initial state: $v_{i} = 0$ , spring stretch $= 2$ m; final state: speed $v$ , spring compression $= 1$ m. $0 + \frac{1}{2} k (2)^{2} = \frac{1}{2} (2) v^{2} + \frac{1}{2} k (1)^{2}$ $2 k = v^{2} + \frac{k}{2} \Rightarrow v^{2} = \frac{3 k}{2} \Rightarrow v = \frac{3 k}{2} m/s$

4.2 When two conservative forces act

Solved Example 3 (Block on incline compressing spring)

A block of mass

m

is placed on a smooth incline of angle

θ

. At the base is a spring of force constant

k

at natural length. The block, initially at distance

ℓ

from the spring, is released from rest. Find the speed when the spring is compressed by

ℓ_{1}

Figure for Solved Example 3: Block of mass

m

on a smooth incline of angle

θ

. It is initially at distance

ℓ

along the slope from a spring of force constant

k

(at natural length). On release, it slides down and compresses the spring.

Solution:

Both gravity and spring force are conservative. Take the final position as gravitational zero. Distance descended along incline $= ℓ + ℓ_{1}$ ; drop in height $= (ℓ + ℓ_{1}) sin θ$ . $K_{i} + U_{i} = K_{f} + U_{f}$ $0 + m g (ℓ + ℓ_{1}) sin θ + 0 = \frac{1}{2} m v^{2} + 0 + \frac{1}{2} k ℓ_{1}^{2}$ $v = \frac{2}{m} [m g (ℓ + ℓ_{1}) sin θ - \frac{1}{2} k ℓ_{1}^{2}]$

Solved Example 4 (Ring on rod, spring anchored to wall)

A light spring of natural length

d

and force constant

k

is fixed to a wall; its free end is attached to a smooth ring of mass

m

that can slide on a vertical rod at distance

d

from the wall. Initially the spring makes

3 7^{\circ}

with the horizontal. Find the ring's speed when the spring becomes horizontal. (

sin 3 7^{\circ} = 3/5

Solution:

Stretched length $ℓ = d / cos 3 7^{\circ} = d / (4/5) = 5 d /4$ . So stretch $y = ℓ - d = d /4$ . Vertical drop of ring $h = ℓ sin 3 7^{\circ} = (5 d /4) (3/5) = 3 d /4$ .

Both gravity and spring are conservative. Taking the final (spring-horizontal) position as gravitational zero: $m g h + \frac{1}{2} k y^{2} = \frac{1}{2} m v^{2}$ $\frac{3}{4} m g d + \frac{1}{2} k (\frac{d}{4})^{2} = \frac{1}{2} m v^{2}$ $v = d \frac{3 g}{2 d} + \frac{k}{16 m}$

4.3 When only a non-conservative force acts

Solved Example 5 (Block stopped by friction)

A block of mass

m

moves with initial speed

v

on a rough horizontal surface with friction coefficient

μ

. Find the distance travelled before it stops.

Solution:

Friction is the only force doing work. Applying $W_{net} = Δ K$ : $- μ m g \cdot ℓ = 0 - \frac{1}{2} m v^{2}$ $ℓ = \frac{v ^{2}}{2 μ g}$

4.4 When conservative and non-conservative forces both act

Solved Example 6 (Particle on curved-flat-curved track)

A particle slides on a track whose central portion BC (length

ℓ = 3 m

) is flat and rough with

μ_{k} = 0.20

; the two curved end portions are smooth. The particle is released at A, at height

h = 1.5 m

above the flat part. Where does it finally come to rest?

Figure for Solved Example 6: Particle released at A slides down the smooth curve, loses energy to friction on the rough flat portion BC (length

ℓ = 3 m

), and finally rests at the midpoint E.

Solution:

All initial mechanical energy $m g h$ is eventually lost to friction on the flat part (curved parts are smooth). If the particle covers total distance $s$ on the flat portion before stopping: $m g h = μ_{k} m g s \Rightarrow s = \frac{h}{μ _{k}} = \frac{1.5}{0.20} = 7.5 m$ Starting from B, the particle goes B $\to$ C ( $3 m$ ), rises on curved part, returns to C, comes back C $\to$ B ( $3 m$ ). Total covered on flat = 6 m; remaining $7.5 - 6 = 1.5 m$ must be covered from B towards C. It rests at the midpoint of BC.

Solved Example 7 (Body hauled up a hill)

A body of mass

m

is slowly hauled up a hill of height

h

and base length

ℓ

by a force always tangent to the trajectory. Friction coefficient with the hill is

μ

. Find the work done by the applied force.

Figure for Solved Example 7: Body of mass

m

hauled slowly up a curved hill by a tangential force

F

. The hill has vertical height

h

, horizontal base length

ℓ

, and surface friction coefficient

μ

. Work done against friction depends only on

ℓ

, not on the exact hill shape.

Solution:

"Slowly" means $Δ K = 0$ . Four forces act: gravity, normal (does no work), friction, applied force. Applying $W_{net} = 0$ : $W_{applied} + W_{gravity} + W_{friction} = 0$ $W_{gravity} = - m g h$ . For friction, on an element $d s$ at slope angle $θ$ , $f = μ m g cos θ$ , and $d W_{f} = - f d s = - μ m g cos θ d s = - μ m g d x$ (since $d s cos θ = d x$ ). Integrating over the full base: $W_{friction} = - μ m g ℓ$ Hence $W_{applied} = m g h + μ m g ℓ$ This shows that the total work against friction depends only on the horizontal projection, not the exact shape of the hill.

4.5 In a non-inertial (accelerating) frame - pseudo force work

Solved Example 8 (Bead on a translating sphere)

A smooth sphere of radius

R

translates in a straight line with constant acceleration

a

. A particle at the top of the sphere is released from rest with respect to the sphere. Find its speed relative to the sphere as a function of angle

θ

from the vertical.

Solution:

Work in the sphere's frame (non-inertial). A pseudo-force $ma$ acts on the particle in the direction opposite to the sphere's acceleration.

As the particle slides through angle $θ$ : horizontal displacement $R sin θ$ , vertical drop $R (1 - cos θ)$ . Work by gravity $= m g R (1 - cos θ)$ ; work by pseudo force $= ma R sin θ$ ; normal reaction does no work. Applying WKE theorem in this frame: $\frac{1}{2} m v^{2} = ma R sin θ + m g R (1 - cos θ)$ $v = 2 R (a sin θ + g - g cos θ)$

5. Vertical Circular Motion

A particle of mass $m$ attached to a light inextensible string of length $R$ (or moving on the inside of a smooth circular loop) is projected horizontally from the lowest point with speed $u$ . Because gravity acts throughout, the speed varies along the loop - and so does the tension in the string (or normal reaction on the loop). This is one of the most-tested applications of the work-energy theorem in JEE and NEET.

Figure 1: Vertical circular motion. At a point where the string makes angle

θ

with the downward vertical, tension

T

acts along the string toward centre

O

, and weight

m g

is resolved into radial

m g cos θ

and tangential

m g sin θ

components.

5.1 Speed and tension at a general angle

Let the particle be at angle $θ$ measured from the downward vertical (bottom of circle). Height above the bottom is $h = R (1 - cos θ)$ . Using energy conservation from bottom (speed $u$ ) to this point (speed $v$ ):

\frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g R (1 - cos θ)

v^{2} = u^{2} - 2 g R (1 - cos θ)

Newton's second law in the radial direction (net inward force provides centripetal acceleration):

T - m g cos θ = \frac{m v ^{2}}{R}

T = \frac{m v ^{2}}{R} + m g cos θ = \frac{m u ^{2}}{R} + m g (3 cos θ - 2)

T (θ) = \frac{m u ^{2}}{R} + 3 m g (cos θ - \frac{2}{3})

5.2 Critical condition for completing the loop (string / hollow loop)

For a string, the constraint is $T \geq 0$ (a string cannot push). Tension is minimum at the top ( $θ = 18 0^{\circ}$ , $cos θ = - 1$ ). Setting $T_{top} = 0$ gives the minimum speed at the top:

\frac{m v _{top}^{2}}{R} = m g \Rightarrow v_{top} = g R

Applying energy conservation from bottom to top (height $2 R$ ):

\frac{1}{2} m u_{m i n}^{2} = \frac{1}{2} m (g R) + m g (2 R)

u_{m i n} = 5 g R

At this critical case, the tension at the bottom is

T_{bottom} = \frac{m u _{m i n}^{2}}{R} + m g = 5 m g + m g = 6 m g

At $θ = 9 0^{\circ}$ (side of the loop): $v_{side}^{2} = 5 g R - 2 g R = 3 g R$ , so $T_{side} = 3 m g$ . At $θ = 18 0^{\circ}$ (top): $T_{top} = 0$ .

Universal result. Regardless of initial speed, the difference in tension between the bottom and top of a vertical circle is always

T_{bottom} - T_{top} = 6 m g

This is a favourite JEE assertion-reason question.

5.3 Three regimes of motion (string)

Figure 2: Three regimes of vertical circular motion. Below

2 g R

the particle oscillates; between

2 g R

and

5 g R

the string goes slack and the particle becomes a projectile; at or above

5 g R

it completes the full loop.

Initial speed $u$	Behaviour
$u \geq 5 g R$	Particle completes the full circle. String stays taut throughout.
$u \leq 2 g R$	Particle never rises above the horizontal diameter; it oscillates about the lowest point (like a pendulum).
$2 g R < u < 5 g R$	Particle rises above the horizontal diameter. String goes slack at some angle $ϕ$ where $T = 0$ , and the particle then undergoes projectile motion inside the loop.

5.4 Slack angle in the intermediate regime

Let $ϕ$ be measured from the upward vertical (from top). At the slack point, $T = 0$ , and Newton's law along the (inward) radius:

m g cos ϕ = \frac{m v ^{2}}{R} \Rightarrow v^{2} = g R cos ϕ

Energy conservation between bottom (speed $u$ ) and this point (height $R + R cos ϕ$ ):

\frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g R (1 + cos ϕ)

Substituting $v^{2} = g R cos ϕ$ :

u^{2} = g R cos ϕ + 2 g R (1 + cos ϕ) = 3 g R cos ϕ + 2 g R

cos ϕ = \frac{u ^{2} - 2 g R}{3 g R}

After the string goes slack, the particle continues as a projectile until it re-enters the taut region or hits the ground.

Solved Example 9 (Pendulum with tension equal to weight)

A heavy particle hangs from a fixed point by an inextensible string of length

ℓ

. It is projected horizontally with speed

g ℓ

from the lowest point. Find the inclination of the string to the vertical when the tension equals the weight of the particle. Also find the speed at that instant.

Solution:

Let the string make angle $θ$ with the downward vertical when $T = m g$ . Radial equation: $T - m g cos θ = \frac{m v ^{2}}{ℓ} \Rightarrow m g (1 - cos θ) = \frac{m v ^{2}}{ℓ}$ $v^{2} = g ℓ (1 - cos θ)$ Energy conservation from bottom to that point (height $ℓ (1 - cos θ)$ ): $\frac{1}{2} m (g ℓ) = \frac{1}{2} m v^{2} + m g ℓ (1 - cos θ)$ $g ℓ = g ℓ (1 - cos θ) + 2 g ℓ (1 - cos θ) = 3 g ℓ (1 - cos θ)$ $cos θ = \frac{2}{3}, v = g ℓ \cdot \frac{1}{3} = \frac{g ℓ}{3}$

5.5 Variants: Rigid rod, hollow tube and outer sphere

Figure 3: A string can only pull, so at the top of the loop it needs

v_{top} \geq g R

to stay taut, giving

u_{m i n} = 5 g R

. A rigid rod can push as well as pull, so it only requires the bob to just reach the top (

v_{top} = 0

), giving

u_{m i n} = 4 g R

(a) Body attached to a rigid rod of length R

A rod can push as well as pull, so there is no lower bound $T \geq 0$ . The only requirement to complete the loop is that the particle just reaches the top with zero speed (or more). Energy conservation:

\frac{1}{2} m u_{m i n}^{2} = 0 + m g (2 R)

u_{m i n} = 4 g R

If $u < 4 g R$ , the rod prevents the particle from leaving the circle; it simply oscillates back and forth.

(b) Body inside a hollow tube or between two rings

A hollow tube (or a double-ring arrangement) can supply normal reaction both inward and outward. This behaves exactly like the rigid rod case: minimum speed to complete a full loop is $u_{m i n} = 4 g R$ .

The angle at which the normal reaction on the body changes direction (from inward to outward, or vice versa) is where $T = 0$ , given by the same slack-angle formula:

cos ϕ = \frac{u ^{2} - 2 g R}{3 g R}

(c) Body sliding on outer surface of a smooth sphere

Figure 4: Body sliding on the outer surface of a smooth sphere. It leaves the surface at point C where

cos ϕ = 2/3

; the height of C above the ground is

\frac{5 R}{3}

At angle $ϕ$ from the top vertical, radial equation (weight radially inward, normal $N$ radially outward, centripetal $= m v^{2} / R$ inward):

m g cos ϕ - N = \frac{m v ^{2}}{R} \Rightarrow N = m (g cos ϕ - \frac{v ^{2}}{R})

The body leaves the surface where $N = 0$ , i.e. $v^{2} = g R cos ϕ$ . Energy conservation from top (rest) to the point at height $R cos ϕ$ below top:

0 = \frac{1}{2} m v^{2} - m g R (1 - cos ϕ) \Rightarrow v^{2} = 2 g R (1 - cos ϕ)

Equating:

g R cos ϕ = 2 g R (1 - cos ϕ) \Rightarrow 3 cos ϕ = 2

cos ϕ = \frac{2}{3}, height above ground = R + R cos ϕ = \frac{5 R}{3}

5.6 Summary of critical speeds and tensions

Setup	Min speed at bottom to complete loop	Key feature
String / open loop (inside)	$5 g R$	String slack at top if $u < 5 g R$
Hollow tube / double ring	$4 g R$	Tube can push and pull
Rigid rod	$4 g R$	Rod can push and pull; oscillation below this speed
Outer sphere	Not applicable	Leaves surface at $cos ϕ = 2/3$

Solved Example 10 (Minimum height for looping)

A small ball slides down from rest on a smooth track and enters a vertical circular loop of radius

R

at the bottom. Find the minimum height

h

(above the bottom of the loop) from which it must be released to complete the loop.

Solution:

At the bottom of the loop the ball needs $u = 5 g R$ . Using conservation of energy from release (rest) to bottom of loop: $m g h = \frac{1}{2} m u^{2} = \frac{1}{2} m (5 g R) \Rightarrow h = \frac{5 R}{2}$

Solved Example 11 (Atwood-like with friction on a table)

In an arrangement, block A (

m_{A} = 4

kg) rests on a rough horizontal table connected via an ideal pulley to block B (

m_{B} = 1

kg) hanging freely. Block A moves twice as fast as B (constraint

v_{A} = 2 v_{B}

). Released from rest, B is found to have speed

0.3

m/s after descending

1 m

. Find the coefficient of friction between A and the table. Take

g = 10 m/s^{2}

Solution:

By constraint, $v_{A} = 0.6$ m/s while $v_{B} = 0.3$ m/s. Also, distance moved by A is $s_{A} = 2 s_{B} = 2 m$ . Apply extended WKE theorem: $W_{n c} = Δ K + Δ U$ Friction on A is the only non-conservative force: $W_{n c} = - μ m_{A} g s_{A} = - μ (4) (10) (2) = - 80 μ$ . $Δ U = - m_{B} g s_{B} = - (1) (10) (1) = - 10 J$ . $Δ K = \frac{1}{2} (4) (0.6)^{2} + \frac{1}{2} (1) (0.3)^{2} = 0.72 + 0.045 = 0.765 J$ . $- 80 μ = 0.765 - 10 = - 9.235 \Rightarrow μ \approx 0.115$

Common Mistakes to Avoid

Watch out

Only conservative forces conserve mechanical energy. If friction, air drag, or a push acts, do not use $K_{i} + U_{i} = K_{f} + U_{f}$ . Use $W_{n c} + W_{ext} = Δ E_{mech}$ instead.
Missing pseudo force work in accelerating frames. In a non-inertial frame, the pseudo-force does work exactly like any real force. Do not forget its contribution when using WKE in that frame.
Wrong critical speed for the loop. $g R$ is the minimum speed at the top, not at the bottom. At the bottom, you need $5 g R$ for a string / open loop.
Applying $u_{m i n} = 5 g R$ to a rod. A rigid rod can push, so its minimum bottom speed is $4 g R$ , not $5 g R$ .
Mixing angle conventions. In the tension formula $T = m u^{2} / R + 3 m g (cos θ - 2/3)$ , $θ$ is measured from the downward vertical. In the slack-angle formula $cos ϕ = (u^{2} - 2 g R) /3 g R$ , $ϕ$ is from the upward vertical. Pick one convention and stick to it, and always check limits ( $θ = 0$ at bottom, $θ = 18 0^{\circ}$ at top).
Assuming $v^{2}$ is halved between bottom and top. Actually $v_{bottom}^{2} - v_{top}^{2} = 4 g R$ , so if $v_{bottom} = 5 g R$ then $v_{top} = g R$ (a factor $5$ difference in speed).
Body on outer sphere: forgetting the height above ground. The fly-off point is at height $R + R cos ϕ = 5 R /3$ above the ground (with sphere resting on the ground), not $R cos ϕ$ .
Direction of friction in a two-body problem. Friction on A from the table opposes A's motion; do not forget its direction when tallying $W_{n c}$ .
Using $W = F d$ when $F$ varies with position or time. Then $W = \int F (x) d x$ or $W = \int F (t) v (t) d t$ ; a bare product will give the wrong energy change.
Confusing "just reaches top" with "just completes loop". For a string, "just completes" means $T = 0$ at the top (with $v_{top} = g R$ ). For a rod, "just reaches" means $v_{top} = 0$ . These give different critical speeds.

Frequently Asked Questions

Q1. What is the work-energy theorem?

It states that the net work done by all forces on a body equals the change in its kinetic energy: $W_{net} = Δ K = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$ . It is a direct consequence of Newton's second law and holds for constant or variable forces, in any type of motion.

Q2. Does the work-energy theorem apply in non-inertial frames?

Yes, provided you include the work done by pseudo-forces. In a frame with acceleration $a_{frame}$ , each mass $m$ experiences a pseudo-force $- m a_{frame}$ ; its work must be added to the real-force work when equating to $Δ K$ measured in that frame.

Q3. When can we use conservation of mechanical energy?

Only when the work done by non-conservative forces (friction, drag) and by external applied forces is zero. In practice, this means smooth surfaces and no external push. If friction or drag acts, use $W_{nc} + W_{ext} = Δ E_{mech}$ instead.

Q4. What is the minimum speed to complete a vertical circular loop with a string?

At the bottom of the loop, the minimum speed is $u_{min} = 5 g R$ . This ensures the tension at the topmost point is zero (with $v_{top} = g R$ ), which is the critical condition since a string cannot push.

Q5. Why is the minimum speed different for a rigid rod?

A rigid rod can exert force in both directions (push and pull), while a string can only pull. For a rod, the only requirement is that the particle just reaches the top, so $u_{min} = 4 g R$ . For a string, the tension at the top must satisfy $T \geq 0$ , giving the stricter $u_{min} = 5 g R$ .

Q6. What is the tension difference between the bottom and top of a vertical circle?

Regardless of the initial speed (as long as the particle completes the loop), the difference in tension is always $T_{bottom} - T_{top} = 6 m g$ . This universal result follows from combining energy conservation with the centripetal force equation.

Q7. Where does a body sliding down the outside of a smooth sphere leave the surface?

It leaves the surface at the angle $ϕ$ from the top vertical where $cos ϕ = 2/3$ . At that point the normal reaction becomes zero. Its height above the ground (with the sphere resting on the ground) is $5 R /3$ .

Q8. How do you handle a body inside a hollow tube on a vertical loop?

A hollow tube can supply normal reaction both inward and outward, so it behaves like a rigid rod for the completing-loop condition: $u_{min} = 4 g R$ . The angle at which the direction of normal reaction switches is $cos ϕ = (u^{2} - 2 g R) / (3 g R)$ , from the upward vertical.

Q9. In the intermediate regime $2 g R < u < 5 g R$ , what happens?

The particle rises above the horizontal diameter. Somewhere between there and the top, the string goes slack ( $T = 0$ ). Beyond that point, the particle undergoes projectile motion inside the loop, following a parabolic trajectory until it re-enters the taut region or hits the ground.

Q10. How is the work-energy theorem different from Newton's second law?

Newton's second law is a vector equation involving instantaneous force and acceleration. The work-energy theorem is a scalar equation involving quantities integrated over a displacement. It converts a vector problem into a scalar one and is especially powerful when force varies with position (springs, gravity in extended motion) or when you only need speed at endpoints (not intermediate accelerations).

Previous year questions on Work Energy Theorem And Its Applications

44 questions from past papers, each with a step-by-step solution.

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1. Statement of the Work-Energy Theorem

2. Derivation

2.1 One-dimensional variable force

2.2 Three-dimensional derivation

3. Extended Form and Conservation of Mechanical Energy

4. Applications of the Work-Energy Theorem

4.1 When only one conservative force acts

4.2 When two conservative forces act

4.3 When only a non-conservative force acts

4.4 When conservative and non-conservative forces both act

4.5 In a non-inertial (accelerating) frame - pseudo force work

5. Vertical Circular Motion

5.1 Speed and tension at a general angle

5.2 Critical condition for completing the loop (string / hollow loop)

5.3 Three regimes of motion (string)

5.4 Slack angle in the intermediate regime

5.5 Variants: Rigid rod, hollow tube and outer sphere

(a) Body attached to a rigid rod of length R

(b) Body inside a hollow tube or between two rings

(c) Body sliding on outer surface of a smooth sphere

5.6 Summary of critical speeds and tensions

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the work-energy theorem?

Q2. Does the work-energy theorem apply in non-inertial frames?

Q3. When can we use conservation of mechanical energy?

Q4. What is the minimum speed to complete a vertical circular loop with a string?

Q5. Why is the minimum speed different for a rigid rod?

Q6. What is the tension difference between the bottom and top of a vertical circle?

Q7. Where does a body sliding down the outside of a smooth sphere leave the surface?

Q8. How do you handle a body inside a hollow tube on a vertical loop?

Q9. In the intermediate regime 2gR​<u<5gR​, what happens?

Q10. How is the work-energy theorem different from Newton's second law?

Previous year questions on Work Energy Theorem And Its Applications

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Q9. In the intermediate regime $2 g R < u < 5 g R$ , what happens?