Work, Energy and Power

Work, Energy and PowerNEET Physics · Class 11 · NCERT Chapter 5

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Work, Energy and Power. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Work calculator (W = F·d·cos θ)

Variable force: work = area under F-x graph

Spring potential energy U = ½ k x²

Free fall: PE → KE conversion

Work-energy theorem checker

Power calculator (P = F·v and P_avg = W/t)

Vertical circular motion (loop)

Work

Work done by a constant force, plus the area-under-F-x picture for variable forces.

Work

Work calculator (W = F·d·cos θ)

Drag F, d and the angle to see the work done by a constant force. Watch the sign change as the angle crosses 90°.

Force F: 20 N

Displacement d: 5 m

Angle θ between F and d: 60°

Work done

W = 50.00 J

W = F d cos θ = 20 \times 5 \times cos 60° = 50.00 J

Force has a component along motion → positive work, body gains KE.

Try this

Set θ = 0 → maximum work F·d. Force entirely along motion.
Set θ = 90 → work is zero. The classic centripetal-force result.
Set θ = 180 → maximum negative work. Friction opposing motion is the canonical case.

Work

Variable force: work = area under F-x graph

Pick a force profile and the final position. The shaded area is the work done by the force from 0 to x.

Slope/value k: 4

Final position x: 5 m

Work = signed area

W = 50.00 J

F = k·x (driving)

Try this

F = kx → linear ramp, area is a triangle: W = ½·xₘₐₓ·F(xₘₐₓ) = ½·k·xₘₐₓ².
F = −kx → restoring force does negative work as x grows; this is exactly minus the spring PE.
Constant force → rectangle, W = F·xₘₐₓ.
Doubling xₘₐₓ for the linear case quadruples W (area scales as x²). Same scaling as spring energy.

Energy and the work-energy theorem

Spring potential energy, the PE-KE conversion in free fall, and the work-energy theorem applied directly.

Spring energy

Spring potential energy U = ½ k x²

Stretch a spring and watch the energy and restoring force update. The square in U = ½ k x² is the most-tested feature on NEET.

Spring constant k: 200 N/m

Extension x: 0.10 m (10 cm)

Stored energy

1.000 J

Restoring force

20.00 N

U = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.10)^{2} = 1.000 J

F = k x = 200 \times 0.10 = 20.00 N

Try this

Double the extension → energy quadruples (square scaling). NEET trap: students often forget the square.
Doubling k at the same x doubles both the force and the energy.
For the same force, a stiffer spring stores less energy (because x = F/k is smaller).

Conservation of energy

Free fall: PE → KE conversion

Watch how kinetic and potential energy trade off during a free fall. Their sum is the total mechanical energy and stays constant.

Drop a ball of mass m from height H. As it falls, gravitational PE converts to KE while the total stays constant. Slide the height to see the energy mix at any instant.

Initial height H: 20 m

Mass m: 1 kg

Current height h: 20.0 m (above ground)

Speed at this height: 0.00 m/s

Energy budget (total = 200.00 J)

PE 100%

Kinetic

0.00 J

Potential

200.00 J

K E + U = const = m g H = 200.00 J

Try this

Slide h all the way to 0 → all energy is KE, ball moves at v = √(2gH).
Slide h to H → all energy is PE, ball at rest momentarily.
Mass cancels: change m and the speed at the bottom stays the same. v = √(2gH) is mass-independent.

Work-energy theorem

Work-energy theorem checker

Pick the initial and final speeds; the widget computes the net work needed to make the change. The sign tells you whether forces accelerate or decelerate the body.

Mass m: 2 kg

Initial speed u: 0 m/s

Final speed v: 10 m/s

Initial KE

0.00 J

Final KE

100.00 J

Net work needed

W_net = 100.00 J

Positive — net force in the direction of motion.

W_{net} = Δ K = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = 100.00 - 0.00 = 100.00 J

Try this

Set u = v → W_net = 0. No net work needed to keep a body at constant speed (though individual forces may do work that cancels).
Triple the speed → ΔKE scales as v² (9× the kinetic energy). Stopping a fast body needs much more work than slowing a slow one.
For a body brought to rest from speed u, W_net = −½mu² (negative, energy dissipated by brakes/friction).

Power and vertical circular motion

Instantaneous and average power, plus the vertical loop where energy conservation meets centripetal force.

Power

Power calculator (instantaneous and average)

Two common NEET setups: instantaneous power P = F·v for a car at constant speed, and average power for a pump lifting water.

Instantaneous: car at constant speed

Driving force F: 500 N

Speed v: 20 m/s

P = F·v

10000 W = 10.00 kW = 13.40 HP

Average: pump lifting water

Mass m: 300 kg

Height h: 20 m

Time t: 60 s

P_avg = mgh / t

1000 W = 1.00 kW

P_{inst} = F v = 500 \times 20 = 10000 W

P_{avg} = \frac{m g h}{t} = \frac{300 \times 10 \times 20}{60} = 1000 W

Try this

For a car at constant speed, the engine must provide power exactly equal to F·v to balance friction.
For a pump, doubling the height doubles the power needed at the same flow rate.
Useful conversion: 1 HP ≈ 746 W. A 100 HP engine puts out ~74.6 kW.

Vertical loop

Vertical circular motion

A mass on a string in a vertical loop. The minimum speed at the bottom (√5gr) is the most-tested NEET fact in this chapter.

Loop radius r: 2 m

Speed at bottom v_bot: 10 m/s

Min v at top

4.47 m/s

Min v at bottom

10.00 m/s

Loop status

Completes the loop. Speed at top = 4.47 m/s

v_{top,min} = g r = 10 \times 2 = 4.47 m/s

v_{bot,min} = 5 g r = 100 = 10.00 m/s

v_{top}^{2} = v_{bot}^{2} - 4 g r = 100 - 80 = 20.00

Try this

Set v_bot just below √(5gr) → loop fails. Slightly above and it just barely makes it.
At v_bot = √(5gr), the speed at the top is exactly √(gr) and tension at the top is zero.
Doubling r requires √2 ≈ 1.41× the speed at the bottom.

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