Class 11 Physics · Original EZT practice
Work, Energy and Power (Class 11 Chapter 5): summary and practice
Work, energy and power give motion a bookkeeping system: a force does work W = Fd cos theta, the work-energy theorem turns net work into a change of kinetic energy, conservative forces store potential energy, and power measures how fast it all happens. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Work and the scalar product
Work is the scalar product of force and displacement: W = Fd cos theta. Force along the motion does full work, force across the motion (theta = 90 degrees) does none, and force against the motion does negative work.
2. Kinetic energy and the work-energy theorem
Kinetic energy is KE = (1/2) m v^2. The work-energy theorem says the net work on a body equals its change in kinetic energy - for a constant force and for a variable force alike (sections 5.5 and 5.6).
3. Potential energy and conservation
Gravity stores potential energy PE = mgh, and a spring stores (1/2) k x^2. For conservative forces the total mechanical energy KE + PE stays constant - a falling body trades height for speed.
4. Power
Power is the rate of doing work: P = W/t, which also equals F v for a steady force. The kilowatt-hour is an energy unit: 1 kWh = 3.6 x 10^6 J.
5. Collisions (section 5.11)
Elastic collisions conserve kinetic energy; inelastic ones do not, and a perfectly inelastic collision leaves the bodies stuck together. In one dimension, conservation of momentum plus the elasticity rule fixes both final velocities.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Compute work at an angle
Choose the method: W = F d cos theta.
EZT practice question
A worker pulls a crate with 60 N at 60 degrees to the horizontal and drags it 5 m. How much work does the pull do?
Show the worked solution
- W = F d cos theta.
- W = 60 x 5 x cos 60 = 300 x 0.5.
Answer: W = 150 J.
Check: Only the along-motion component (60 cos 60 = 30 N) works over 5 m: 30 x 5 = 150 J.
Watch for this: Read theta as the angle between force and displacement - picking sin instead of cos is the classic slip.
Now try it yourself
A 40 N force pushes a box 3 m straight ahead. Work done?
Check your answer
120 J.
cos 0 = 1, so W = 40 x 3 = 120 J.
Find kinetic energy
Choose the method: KE = (1/2) m v^2.
EZT practice question
What is the kinetic energy of a 2 kg ball moving at 3 m/s?
Show the worked solution
- KE = (1/2) x 2 x 3^2.
- KE = 1 x 9.
Answer: KE = 9 J.
Check: Doubling speed would give 36 J - four times, as v^2 demands.
Watch for this: Do not drop the 1/2, and square the speed before multiplying by mass.
Now try it yourself
Find the kinetic energy of a 0.5 kg stone at 4 m/s.
Check your answer
4 J.
KE = 0.5 x 0.5 x 16 = 4 J.
Use the work-energy theorem
Choose the method: Net work equals the change in kinetic energy.
EZT practice question
A 1500 kg car accelerates from rest to 20 m/s. How much net work is done on it?
Show the worked solution
- Net work = change in KE = (1/2) m v^2 - 0.
- W = 0.5 x 1500 x 400.
Answer: W = 300000 J (3 x 10^5 J).
Check: No force or distance was needed - the theorem jumps straight to the energy change.
Watch for this: Use the CHANGE in KE; starting from rest makes the initial term zero, but that is luck, not a rule.
Now try it yourself
The same car brakes from 20 m/s to rest. Net work done on it?
Check your answer
-300000 J.
Final KE is zero, so net work = 0 - 300000 J: negative work removes energy.
Trade height for speed
Choose the method: mgh lost becomes (1/2) m v^2 gained, so v = sqrt(2gh).
EZT practice question
A 2 kg stone falls 5 m from rest with no air resistance (g = 9.8 m/s^2). Find the potential energy lost and the speed at the bottom.
Show the worked solution
- PE lost = mgh = 2 x 9.8 x 5 = 98 J.
- 98 = (1/2) x 2 x v^2, so v = sqrt(98).
- sqrt(98) is about 9.9.
Answer: 98 J lost, and v is about 9.9 m/s at the bottom.
Check: v = sqrt(2gh) = sqrt(2 x 9.8 x 5) = sqrt(98) - same result, mass cancels.
Watch for this: Do not use v = gt unless you also find the time first - energy gets there in one line.
Now try it yourself
A 1 kg ball drops 10 m. Speed at the bottom?
Check your answer
14 m/s.
v = sqrt(2 x 9.8 x 10) = sqrt(196) = 14 m/s.
Measure power
Choose the method: P = W/t (and P = F v at steady speed).
EZT practice question
A motor does 2000 J of work in 10 s. What is its power?
Show the worked solution
- P = W/t.
- P = 2000/10.
Answer: P = 200 W.
Check: At 200 W the motor delivers 200 J every second, so 10 s gives 2000 J.
Watch for this: Power cares about time - the same work done faster means more power.
Now try it yourself
A lifter does 100 J of work in 20 s. Power?
Check your answer
5 W.
P = 100/20 = 5 W.
Convert kilowatt-hours
Choose the method: 1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J.
EZT practice question
A 1500 W heater runs for 1 hour. How much energy does it use, in kWh and in joules?
Show the worked solution
- Energy = power x time = 1.5 kW x 1 h = 1.5 kWh.
- In joules: 1500 x 3600 = 5.4 x 10^6 J.
Answer: 1.5 kWh, which is 5.4 x 10^6 J.
Check: 1.5 x 3.6 x 10^6 = 5.4 x 10^6 - consistent with the conversion.
Watch for this: kWh is energy, not power - a 2 kW heater used for 3 h consumes 6 kWh.
Now try it yourself
A 100 W bulb glows for 10 h. Energy used?
Check your answer
1 kWh (3.6 x 10^6 J).
0.1 kW x 10 h = 1 kWh.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Work formula?
Show answer and reason
W = F d cos theta.
Only the force component along the displacement does work.
2. Kinetic energy?
Show answer and reason
KE = (1/2) m v^2.
Net work changes KE by exactly its own amount.
3. Gravitational potential energy?
Show answer and reason
PE = mgh.
Height stores energy against gravity.
4. 1 kWh in joules?
Show answer and reason
3.6 x 10^6 J.
1000 W for 3600 s.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A 40 N force pushes a box 3 m in its own direction. Find the work done.
- Find the kinetic energy of a 0.5 kg stone moving at 4 m/s.
- A device does 100 J of work in 20 s. Find its power.
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Questions about this practice page
- What is the work done by a force at 90 degrees to the motion?
- Zero. W = Fd cos theta and cos 90 = 0, so a perpendicular force - like the string tension in uniform circular motion - does no work at all.
- What does the work-energy theorem say?
- The net work done on a body equals its change in kinetic energy. It holds for constant forces and for variable forces, which the chapter proves in sections 5.5 and 5.6.
- How do elastic and inelastic collisions differ?
- As section 5.11 describes them: an elastic collision conserves kinetic energy, while an inelastic one loses some to heat, sound or deformation. Momentum is conserved in both, as long as no net external force acts.
- Is a kilowatt-hour a unit of power?
- No - it is energy: one kilowatt running for one hour, which is 3.6 x 10^6 J. Electricity bills charge for kWh, the energy used, not the kW rating.
- Are these the NCERT exercise questions for Work, Energy and Power (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.