Chapter 6: Work Energy and Power (Class 11)

1. When is work said to be done by a force?

When the force produces a displacement in the object along its line of action.

2. What is the SI unit of work?

joule (J).

3. What is the work done when force is perpendicular to displacement?

Zero (as cos 90° = 0).

4. Can work done be negative? Give an example.

Yes. Work done by friction when an object slides.

5. Define Kinetic Energy (KE).

Energy possessed by a body due to its motion.

6. Write the formula for Kinetic Energy.

K = ½mv².

7. State the Work-Energy Theorem.

Net work done on a body equals the change in its kinetic energy (W_net = ΔK).

8. Define Power.

The rate of doing work or transferring energy (P = W/t).

9. What is the SI unit of Power?

watt (W). (1 W = 1 J/s).

10. Define Potential Energy (PE).

Energy possessed by a body due to its position or configuration.

11. Write the formula for gravitational potential energy near Earth’s surface.

U = mgh.

12. What are conservative forces?

Forces for which work done depends only on initial and final positions, not path. (e.g., gravity, spring force).

13. Give an example of a non-conservative force.

Friction, Air drag.

14. State the Law of Conservation of Mechanical Energy.

If only conservative forces act, total mechanical energy (K+U) remains constant.

15. In which type of collision is kinetic energy conserved?

Elastic collision.

16. What happens in a perfectly inelastic collision?

The bodies stick together after collision.

17. What quantity is always conserved in all types of collisions (isolated system)?

Linear Momentum.

18. What is the value of coefficient of restitution ‘e’ for a perfectly elastic collision?

e = 1.

19. What is the minimum speed required at the top for an object to complete a vertical circle on a string?

v_top = √(gR).

20. What is the dimensional formula for Power?

[ML²T⁻³].

1. Explain positive, negative, and zero work done with examples.

• Positive Work (+W): Force component along displacement (θ < 90°). Ex: Lifting an object (work by lifting force).
• Negative Work (-W): Force component opposite to displacement (θ > 90°). Ex: Work done by friction on moving object.
• Zero Work (W=0): Force perpendicular to displacement (θ = 90°) OR displacement is zero. Ex: Work done by gravity on object moving horizontally.

2. State and explain the work-energy theorem.

• Statement: Net work done (W_net) by all forces acting on a body equals the change in its kinetic energy (ΔK).
• W_net = K_final – K_initial = ½mv² – ½mu².
• Explanation: It connects work (cause of change in motion) to kinetic energy (energy of motion). If positive net work is done, KE increases; if negative, KE decreases.

3. Differentiate between conservative and non-conservative forces.

• Conservative: Work done independent of path, depends only on endpoints. Work over closed loop is zero. Ex: Gravity, Spring Force. Potential energy defined.
• Non-Conservative: Work done depends on the path taken. Work over closed loop is non-zero (usually negative, energy dissipated). Ex: Friction, Air Drag.

4. Write the relation between Kinetic Energy (K) and Linear Momentum (p).

• K = ½mv² and p = mv.
• Substitute v = p/m into K: K = ½ m (p/m)² = p²/2m.
• Also, p = √(2mK).

5. Define average power and instantaneous power.

• Average Power (P_avg) = Total Work (W) / Total Time (t).
• Instantaneous Power (P) = Rate of doing work at an instant = dW/dt. Also P = F ⋅ v.

6. What is potential energy of a spring? On what factors does it depend?

• Elastic potential energy stored in a spring when compressed or extended.
• Formula: U = ½kx², where ‘k’ is spring constant (stiffness) and ‘x’ is displacement from equilibrium.
• Depends on stiffness (k) and amount of compression/extension (x²).

7. When is total mechanical energy conserved? State the law.

• Total mechanical energy (E = K + U) is conserved when ONLY conservative forces are acting on the system.
• Law: If net work done by non-conservative forces is zero, the total mechanical energy of the system remains constant (ΔE = ΔK + ΔU = 0).

8. Differentiate between elastic and inelastic collisions.

• Elastic: BOTH Linear Momentum AND Kinetic Energy are conserved. (e=1).
• Inelastic: Linear Momentum is conserved, BUT Kinetic Energy is NOT conserved (some KE lost). (0 ≤ e < 1).
• Perfectly Inelastic: Momentum conserved, Max KE lost, bodies stick together (e=0).

9. Why is work done by centripetal force always zero?

• Centripetal force acts towards the center of the circular path.
• Instantaneous displacement (or velocity) of the object is always tangent to the circle.
• The angle (θ) between the centripetal force (inward radius) and the displacement/velocity (tangent) is always 90°.
• Work Done W = Fd cos θ = F_c d cos 90° = F_c d (0) = 0.

10. Can kinetic energy be negative? Can potential energy be negative?

• Kinetic Energy (K = ½mv²): Mass (m) is positive. Speed squared (v²) is always positive or zero. Therefore, KE is always **positive or zero**. It cannot be negative.
• Potential Energy (U): PE depends on the choice of reference level (where U=0). It can be **positive, negative, or zero** depending on the position relative to the reference level. Ex: Gravitational PE can be negative if object is below reference level.

11. Calculate work done in lifting a 5 kg box vertically by 2 metres. (g=10 m/s²).

• Force needed (at least) = Weight = mg = 5 kg × 10 m/s² = 50 N.
• Displacement d = 2 m (upwards).
• Force and displacement are in same direction (θ=0°).
• Work W = Fd cos 0° = 50 N × 2 m × 1 = 100 J.
• (Alternatively, Work = Change in PE = mg(h_f-h_i) = 5×10×(2-0) = 100 J).

12. Define coefficient of restitution (e).

• ‘e’ measures elasticity of a collision (for head-on collision).
• Defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision.
• e = | v₂ - v₁ | / | u₁ - u₂ |.
• e=1 (Elastic), e=0 (Perfectly Inelastic), 0

13. A light body and a heavy body have same momentum. Which has greater KE?

• KE = p²/2m. Momentum (p) is same for both.
• KE is inversely proportional to mass (m).
• Therefore, the lighter body (smaller m) will have greater Kinetic Energy.

14. A light body and a heavy body have same KE. Which has greater momentum?

• p = √(2mK). Kinetic Energy (K) is same.
• Momentum (p) is directly proportional to the square root of mass (√m).
• Therefore, the heavier body (larger m) will have greater Momentum.

15. Give an example of motion in a vertical circle.

• Swinging a bucket of water in a vertical loop.
• A giant wheel ride (ferris wheel) at constant angular speed (motion of a person).
• A stone tied to a string whirled vertically.

16. Is work a scalar or a vector? Why?

• Work is a Scalar quantity.
• Reason: It is defined as the scalar (dot) product of two vector quantities (Force F and Displacement d).
• W = F ⋅ d = Fd cos θ. The result of a dot product is always a scalar.

1. State and prove the Work-Energy Theorem for a variable force.

Statement: Net work done by all forces equals change in kinetic energy (W_net = ΔK).

Proof for Variable Force (1D):

• From Newton’s 2nd Law: F_net = ma = m(dv/dt).
• Using chain rule: F_net = m (dv/dx) (dx/dt) = m (dv/dx) v = mv (dv/dx).
• So, F_net dx = mv dv.
• Work done (dW) for small displacement dx is dW = F_net dx = mv dv.
• Total work W_net in changing velocity from u to v (position xᵢ to x_f):
• Wnet = ∫xᵢxf Fnet dx = ∫uv mv dv
• Integrate RHS: ∫uv mv dv = m [v²/2]uv = m (v²/2 - u²/2)
• Wnet = ½mv² - ½mu² = Kfinal - Kinitial = ΔK.
• Hence Proved.

2. Define potential energy. Derive the expression for gravitational potential energy near the Earth’s surface and elastic potential energy of a spring.

Potential Energy (U): Energy stored by virtue of position or configuration.

Gravitational PE (Near Surface):

• Consider lifting mass ‘m’ vertically up by height ‘h’ against gravity (force mg downwards).
• Work done by lifting force = Force × displacement = (mg) × h = mgh.
• This work is done against the conservative gravitational force.
• Change in PE (ΔU) = – (Work done by conservative force) = – (Work done by gravity).
• Work by gravity = Force × displacement × cos 180° = (mg)(h)(-1) = -mgh.
• ΔU = U_final – U_initial = – (-mgh) = mgh.
• If we choose reference U_initial = 0 at h=0, then U_final = U = mgh.
• U = mgh.

Elastic PE of Spring:

• Consider stretching/compressing spring (constant k) by distance x from equilibrium.
• Restoring Force F_s = -kx (variable force, opposes displacement).
• Work done by external force (F_ext = +kx) to stretch from 0 to x:
• Wext = ∫0x Fext dx = ∫0x kx dx
• Wext = k [x²/2]0x = k(x²/2 - 0²/2) = ½kx².
• This work is stored as Elastic Potential Energy (U_s) because spring force is conservative.
• U_s = ½kx².

3. State and prove the Law of Conservation of Mechanical Energy for a freely falling body.

Statement: If only conservative forces (like gravity) act, total mechanical energy (E = K + U) is constant.

Proof for Freely Falling Body:

• Consider body (mass m) falling from height h (Point A). Reference U=0 at ground.
• Point A (Top, height h): Speed u=0. K_A = 0. U_A = mgh. Total Energy E_A = K_A + U_A = 0 + mgh = mgh.
• Point B (Intermediate, height y): Distance fallen = h-y. Velocity v_B. Use v²=u²+2as: v_B² = 0² + 2g(h-y).
  K_B = ½mv_B² = ½m[2g(h-y)] = mg(h-y). U_B = mgy.
  Total Energy E_B = K_B + U_B = mg(h-y) + mgy = mgh – mgy + mgy = mgh.
• Point C (Ground, height 0): Distance fallen = h. Velocity v_C. Use v²=u²+2as: v_C² = 0² + 2gh.
  K_C = ½mv_C² = ½m(2gh) = mgh. U_C = mg(0) = 0.
  Total Energy E_C = K_C + U_C = mgh + 0 = mgh.
• Since E_A = E_B = E_C = mgh (constant), mechanical energy is conserved.

4. Differentiate between Elastic and Inelastic collisions with examples. Define coefficient of restitution.

Collisions involve interaction between bodies for a short time.

Elastic Collision:

• Total Linear Momentum is conserved.
• Total Kinetic Energy is conserved.
• Forces involved are conservative.
• Objects regain original shape after impact (ideally).
• Example: Collisions between subatomic particles, nearly elastic collision of billiard balls.
• Coefficient of Restitution (e) = 1.

Inelastic Collision:

• Total Linear Momentum is conserved.
• Total Kinetic Energy is NOT conserved (some lost as heat, sound, deformation).
• May involve non-conservative forces during impact.
• Objects may deform permanently.
• Example: Car crash, ball hitting mud, most real-life collisions.
• Coefficient of Restitution (e): 0 ≤ e < 1. (e=0 for perfectly inelastic where bodies stick together).

Coefficient of Restitution (e):

• Defined as ratio of relative velocity of separation to relative velocity of approach along the line of impact.
• e = | v₂ - v₁ | / | u₁ - u₂ | (for 1D collision).

5. Derive the expression for maximum speed of a vehicle on a level circular road.

• Car (mass m), speed v, radius r, coeff. static friction μ_s.
• Required inward force = Centripetal force = mv²/r.
• This force is provided *only* by static friction (f_s) between tyres and road, directed towards the center.
• Maximum available static friction f_s(max) = μ_sN.
• On a level road, Normal Reaction N = Weight mg.
• So, f_s(max) = μ_smg.
• For safe turn (no slipping), Required Force ≤ Available Force:
• mv²/r ≤ μ_smg
• v²/r ≤ μ_sg
• v² ≤ μ_srg
• The maximum speed occurs when equality holds.
• v_max = √(μ_srg).

6. Derive the relation P = Fv for instantaneous power.

• Instantaneous power (P) is the rate of doing work: P = dW/dt.
• Work done (dW) by a force F for a small displacement dr is dW = F ⋅ dr.
• Substitute this into the power definition: P = d/dt (F ⋅ dr).
• If the force F is constant over the small time dt: P = F ⋅ (dr/dt).
• We know that instantaneous velocity v = dr/dt.
• Therefore, P = F ⋅ v.
• In magnitude, if θ is the angle between F and v, P = Fv cos θ.

7. Discuss the motion of a body in a vertical circle, deriving the minimum speeds required at the top and bottom to complete the loop (mass on a string).

Consider mass ‘m’ tied to string of length ‘R’, whirled vertically.

• Forces acting: Weight (mg) downwards, Tension (T) along string towards center.
• Net force towards center provides centripetal force (mv²/R).

At Topmost Point (T):

• Both T and mg act downwards (towards center).
• Net force = T_T + mg = mv_T²/R.
• For minimum speed to just complete loop, tension can become zero (T_T ≈ 0).
• So, mg = mv_T²/R ⇒ v_T(min) = √(gR).

At Bottommost Point (B):

• Tension T_B acts upwards, mg downwards.
• Net force towards center = T_B – mg = mv_B²/R.
• Energy Conservation (From Top (T) to Bottom (B)): (KE + PE)_T = (KE + PE)_B. (Take PE=0 at B).
• (½mv_T²) + (mg(2R)) = (½mv_B²) + 0.
• Substitute v_T = √(gR): ½m(gR) + 2mgR = ½mv_B².
• mgR(½ + 2) = ½mv_B² ⇒ mgR(5/2) = ½mv_B².
• Cancel m, multiply by 2: 5gR = v_B².
• v_B(min) = √(5gR).

8. Define Conservative Force. Show that gravitational force is a conservative force.

Conservative Force: Work done by this force in moving a particle between two points depends only on the initial and final points, not on the path taken. Equivalently, work done over any closed path is zero.

Proof for Gravitational Force (Near Earth):

• Consider moving mass ‘m’ from A(x₁, y₁) to B(x₂, y₂) near Earth (gravity F_g = -mgĵ, taking upward y as positive).
• Path 1: Move horizontally A(x₁,y₁) → C(x₂,y₁) then vertically C(x₂,y₁) → B(x₂,y₂).
  • Work(A→C): F_g is vertical, displacement (x₂-x₁) is horizontal. θ=90°. W_AC = 0.
  • Work(C→B): F_g (-mg) is opposite to displacement (y₂-y₁). W_CB = (-mg)(y₂-y₁)cos 0° ? No. Force and disp in y dir. Force=-mg, Disp = y₂-y₁. W_CB = (-mg)(y₂-y₁).
  • W₁ = W_AC + W_CB = -mg(y₂-y₁).
• Path 2: Move vertically A(x₁,y₁) → D(x₁,y₂) then horizontally D(x₁,y₂) → B(x₂,y₂).
  • Work(A→D): Force=-mg, Disp=y₂-y₁. W_AD = (-mg)(y₂-y₁).
  • Work(D→B): Force vertical, Disp horizontal. θ=90°. W_DB = 0.
  • W₂ = W_AD + W_DB = -mg(y₂-y₁).
• Since W₁ = W₂ (depends only on initial y₁ and final y₂), work is path independent.
• Work over closed loop A→C→B→D→A = W₁ + W₂’ = W₁ – W₂ = 0.

Therefore, gravitational force (near Earth) is conservative.

9. Explain elastic and inelastic collisions in two dimensions (Oblique Collisions).

In 2D collisions, objects move in a plane after impact.

• Conservation of Momentum: Total linear momentum is conserved in **both x and y directions separately** (assuming no external force).
  • Σp_{initial, x} = Σp_{final, x}
  • Σp_{initial, y} = Σp_{final, y}
• Elastic Collision (2D):
  • Momentum conserved along x and y axes.
  • Kinetic Energy is also conserved: ΣK_initial = ΣK_final.
  • These give three equations relating initial and final velocities.
• Inelastic Collision (2D):
  • Momentum conserved along x and y axes.
  • Kinetic Energy is NOT conserved (K_final < K_initial).
  • If perfectly inelastic, bodies stick together: v_1f = v_2f (both in x and y components).
• Solving 2D collision problems usually involves resolving velocities into x and y components before applying conservation laws.

10. Show that in a perfectly elastic head-on collision between two bodies of equal masses, their velocities are exchanged after collision.

Consider two bodies (mass m) with initial velocities u₁ and u₂. After elastic head-on collision, final velocities are v₁ and v₂.

• 1. Conservation of Momentum:
  mu₁ + mu₂ = mv₁ + mv₂
  Divide by m: u₁ + u₂ = v₁ + v₂
  ⇒ v₂ – v₁ = u₁ – u₂ — (Eq 1)
• 2. Conservation of Kinetic Energy:
  ½mu₁² + ½mu₂² = ½mv₁² + ½mv₂²
  Divide by ½m: u₁² + u₂² = v₁² + v₂²
  Rearrange: u₁² – v₁² = v₂² – u₂²
  Factorize: (u₁ – v₁)(u₁ + v₁) = (v₂ – u₂)(v₂ + u₂) — (Eq 2)
• 3. Divide Eq 2 by Eq 1:
  [(u₁ – v₁)(u₁ + v₁)] / [u₁ – v₁] = [(v₂ – u₂)(v₂ + u₂)] / [v₂ – u₂] *Wait, easier to use e=1 definition.*
• Alternate (Using Coefficient of Restitution e=1):
  e = (Relative Velocity of Separation) / (Relative Velocity of Approach) = 1.
  (v₂ - v₁) / (u₁ - u₂) = 1
  ⇒ v₂ – v₁ = u₁ – u₂ — (Eq A – same as Eq 1 from momentum)
• From Momentum Conservation:
  u₁ + u₂ = v₁ + v₂ — (Eq B)
• Solve A and B: Add Eq A and Eq B:
  (v₂ – v₁) + (v₁ + v₂) = (u₁ – u₂) + (u₁ + u₂)
  2v₂ = 2u₁ ⇒ v₂ = u₁
• Substitute v₂ = u₁ into Eq B:
  u₁ + u₂ = v₁ + u₁ ⇒ v₁ = u₂

Conclusion: Final velocity of first body (v₁) equals initial velocity of second (u₂), and final velocity of second body (v₂) equals initial velocity of first (u₁). Their velocities are exchanged.

11. Prove Work-Energy theorem for a constant force.

Assume a constant force F acts on a body of mass m, causing displacement s along the direction of force, changing its velocity from u to v.

• Since force is constant, acceleration ‘a’ is constant. a = F/m.
• Use the 3rd equation of motion: v² = u² + 2as.
• Rearrange for ‘as’: as = (v² – u²) / 2.
• Work done by the constant force W = Force × Displacement = F × s.
• Substitute F = ma: W = (ma) × s = m(as).
• Substitute the value of ‘as’ from the equation of motion:
• W = m [ (v² – u²) / 2 ]
• W = ½m(v² – u²) = ½mv² – ½mu².
• W = K_final – K_initial = ΔK.

Thus, work done by the constant net force equals the change in kinetic energy.

12. Explain the concept of Power and its different units.

Power (P):

• Power is the rate at which work is done or energy is transferred. It measures how fast work is being performed. (Yeh batata hai ki kaam kitni tezi se ho raha hai).
• Average Power P_avg = Work / Time = W/t.
• Instantaneous Power P = dW/dt = F ⋅ v.
• It is a scalar quantity.

Units of Power:

• SI Unit: watt (W). 1 watt = 1 joule per second (1 W = 1 J/s).
• CGS Unit: erg per second (erg/s). (1 W = 10⁷ erg/s).
• Practical/Commercial Unit: horsepower (hp). Used commonly for engines/motors.
  • 1 hp (British) ≈ 746 W.
  • 1 hp (Metric) ≈ 735.5 W.
• Dimensional Formula: [ML²T⁻³].

Note: Kilowatt-hour (kWh) is a unit of *Energy*, not Power. 1 kWh = (1000 W) × (3600 s) = 3.6 × 10⁶ J.

13. Define Elastic Potential Energy. A spring of force constant ‘k’ is stretched by ‘x’. Calculate the work done and the energy stored.

Elastic Potential Energy (U_s): Energy stored in a body due to its change in shape or configuration (deformation), like in a stretched or compressed spring.

Work Done & Energy Stored:

• Spring Force F_s = -kx (Restoring force, opposite to displacement x from equilibrium).
• External Force needed to stretch slowly F_ext = +kx.
• Work done by *external* force in stretching from 0 to x is:
• Wext = ∫0x Fext dx = ∫0x kx dx = k[x²/2]0x = ½kx².
• This work done by the external force is stored as Elastic Potential Energy in the spring (as spring force is conservative).
• Stored Energy U_s = ½kx².
• Work done *by the spring force* during stretching is W_s = -W_ext = -½kx². (Negative work as force opposes displacement).

14. What is the difference between conservative and non-conservative forces regarding energy?

• Conservative Forces:
  • Work done is path independent. Work in closed loop = 0.
  • **Mechanical energy (KE + PE) is conserved** when only these forces act.
  • Potential energy can be defined for them.
  • Ex: Gravity, Electrostatic force, Spring force.
• Non-Conservative Forces:
  • Work done depends on the path taken. Work in closed loop ≠ 0.
  • **Mechanical energy is NOT conserved** when these forces act. Energy is usually dissipated as heat or other forms.
  • Potential energy is not defined for them in the same way.
  • Ex: Friction, Air drag, Viscous force.

15. Two bodies of masses m₁ and m₂ undergo a perfectly inelastic head-on collision. Find the loss in kinetic energy.

Let initial velocities be u₁ and u₂. In perfectly inelastic collision, they stick together and move with a common final velocity V.

• 1. Find Common Velocity (V): Use Conservation of Momentum.
  m₁u₁ + m₂u₂ = (m₁ + m₂)V
  ⇒ V = (m₁u₁ + m₂u₂) / (m₁ + m₂).
• 2. Initial Kinetic Energy (K_i):
  K_i = ½m₁u₁² + ½m₂u₂².
• 3. Final Kinetic Energy (K_f):
  Both masses move with velocity V.
  K_f = ½(m₁ + m₂)V².
• 4. Loss in Kinetic Energy (ΔK):
  ΔK = K_i – K_f.
  Substitute V: ΔK = (½m₁u₁² + ½m₂u₂²) – ½(m₁ + m₂)[ (m₁u₁ + m₂u₂)/(m₁ + m₂) ]².
• Simplify: ΔK = (½m₁u₁² + ½m₂u₂²) – ½ (m₁u₁ + m₂u₂)² / (m₁ + m₂).
• Further simplification (takes algebraic steps, using relative velocity urel = u₁-u₂ leads to):
  ΔK_loss = ½ [ m₁m₂ / (m₁ + m₂) ] (u₁ – u₂)².

This shows KE loss is always positive (or zero if velocities were initially same) and depends on masses and square of relative velocity of approach.