Chapter 5: Work, Energy and Power
Unit IV – Work, Energy and Power
5.1 Work Done by a Force
By a Constant Force
\[ W = \vec{F} \cdot \vec{s} = Fs\cos\theta \]
where \(\theta\) is the angle between force \(\vec{F}\) and displacement \(\vec{s}\).
- Work is a scalar quantity.
- SI unit: Joule (J) = N·m = kg·m²·s⁻²
- Work is positive when \(\theta \lt 90°\), negative when \(\theta \gt 90°\), zero when \(\theta = 90°\).
By a Variable Force
\[ W = \int_{x_i}^{x_f} F,dx \]
(area under F-x graph)
5.2 Work-Energy Theorem
The net work done on a body equals its change in kinetic energy:
\[ W_{net} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]
Proof: From \(v^2 = u^2 + 2as\):
\[ \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mas = Fs = W \]
5.3 Kinetic Energy and Potential Energy
Kinetic Energy
\[ KE = \frac{1}{2}mv^2 \]
Gravitational Potential Energy
\[ PE = mgh \]
(taking ground as reference)
Spring Potential Energy
A spring compressed/stretched by \(x\) (spring constant \(k\)):
\[ F_{spring} = -kx \quad \text{(Hooke’s Law)} \]
\[ PE_{spring} = \frac{1}{2}kx^2 \]
5.4 Conservative and Non-Conservative Forces
| Conservative Forces | Non-Conservative Forces |
|---|---|
| Work done is path-independent | Work done depends on path |
| Mechanical energy is conserved | Energy is dissipated |
| e.g., gravity, spring force, electrostatic | e.g., friction, air drag, viscosity |
Mathematically, for conservative force: \(\oint \vec{F}\cdot d\vec{r} = 0\)
5.5 Conservation of Mechanical Energy
Total mechanical energy (KE + PE) is conserved in the presence of only conservative forces:
\[ E = KE + PE = \text{constant} \]
At height h (ball falling from height H):
\[ \frac{1}{2}mv^2 + mgh = mgH \]
\[ v = \sqrt{2g(H-h)} \]
Energy Conservation – Falling Ball Graph
5.6 Power
Power is the rate of doing work:
\[ P = \frac{W}{t} = \frac{dW}{dt} = \vec{F}\cdot\vec{v} \]
- SI unit: Watt (W) = J/s
- 1 horsepower (hp) = 746 W
5.7 Motion in a Vertical Circle
For a particle of mass \(m\) moving on a vertical circle of radius \(r\) (on the inside of a loop):
At the bottom:
\[ T_{bottom} - mg = \frac{mv^2_{bottom}}{r} \]
At the top:
\[ T_{top} + mg = \frac{mv^2_{top}}{r} \]
Minimum speed at top (for T = 0):
\[ v_{top,min} = \sqrt{gr} \]
Minimum speed at bottom (to complete the loop):
\[ v_{bottom,min} = \sqrt{5gr} \]
5.8 Elastic and Inelastic Collisions
| Property | Elastic Collision | Perfectly Inelastic |
|---|---|---|
| Momentum | Conserved | Conserved |
| Kinetic Energy | Conserved | Not conserved |
| Bodies after | Separate | Move together |
For 1D elastic collision (\(m_1\) hits stationary \(m_2\)):
\[ v_1’ = \frac{(m_1 - m_2)}{(m_1 + m_2)},u_1 \]
\[ v_2’ = \frac{2m_1}{(m_1 + m_2)},u_1 \]
Special case: If \(m_1 = m_2\): \(v_1’ = 0\), \(v_2’ = u_1\) (velocities exchange)
Coefficient of Restitution:
\[ e = \frac{\text{relative speed after}}{\text{relative speed before}} = \frac{v_2’ - v_1’}{u_1 - u_2} \]
(elastic: \(e = 1\), perfectly inelastic: \(e = 0\))
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(W = Fs\cos\theta\) | Work done |
| \(KE = \frac{1}{2}mv^2\) | Kinetic energy |
| \(PE = mgh\) | Gravitational PE |
| \(PE_{spring} = \frac{1}{2}kx^2\) | Spring PE |
| \(P = Fv\cos\theta\) | Power |
| \(v_{min,top} = \sqrt{gr}\) | Min speed at top of loop |
| \(v_{min,bottom} = \sqrt{5gr}\) | Min speed at bottom of loop |
Practice Questions
Section A – MCQ (1 mark each)
Q1. A body of mass 1 kg is thrown upward with velocity 20 m/s. Its kinetic energy at launch is:
(a) 100 J (b) 200 J (c) 400 J (d) 800 J
Answer
(b) 200 J — \(KE = \frac{1}{2} \times 1 \times 20^2 = 200 \text{ J}\)
Q2. A spring constant is 100 N/m. The energy stored when compressed by 20 cm is:
(a) 1 J (b) 2 J (c) 4 J (d) 10 J
Answer
(b) 2 J — \(PE = \frac{1}{2} \times 100 \times (0.2)^2 = 2 \text{ J}\)
Q3. The work done by a force at 90° to displacement is:
(a) Maximum (b) Minimum (c) Zero (d) Negative
Answer
(c) Zero — \(W = Fs\cos 90° = 0\)
Section B – Short Answer (2–3 marks)
Q4. A car of mass 1000 kg moves with a velocity of 20 m/s. Brakes apply a retarding force of 5000 N. Find: (i) Initial KE (ii) Stopping distance.
Answer
(i) \(KE = \frac{1}{2} \times 1000 \times 400 = 200000 \text{ J} = \mathbf{2 \times 10^5 \text{ J}}\)
(ii) \(W = -F \times d = -\Delta KE\)
\(5000 \times d = 200000 \Rightarrow d = \mathbf{40 \text{ m}}\)
Q5. What is the minimum speed needed at the top of a vertical circular loop of radius 5 m? Also find the minimum speed at the bottom. (\(g = 10 \text{ m/s}^2\))
Answer
\(v_{top} = \sqrt{gr} = \sqrt{10 \times 5} = \mathbf{\sqrt{50} \approx 7.07 \text{ m/s}}\)
\(v_{bottom} = \sqrt{5gr} = \sqrt{5 \times 10 \times 5} = \sqrt{250} \approx \mathbf{15.8 \text{ m/s}}\)
Section D – Competency-Based Questions
Q6. (Case Study) A ball of mass 0.5 kg is dropped from a height of 20 m.
(i) Find its total mechanical energy at the top.
(ii) Find its KE when it has fallen 12 m.
(iii) At what height is KE = PE?
(iv) Find its velocity just before hitting the ground.
(\(g = 10 \text{ m/s}^2\))
Answer
(i) \(E = mgh = 0.5 \times 10 \times 20 = \mathbf{100 \text{ J}}\)
(ii) \(PE\) at 8 m height = \(0.5 \times 10 \times 8 = 40 \text{ J}\)
\(KE = E - PE = 100 - 40 = \mathbf{60 \text{ J}}\)
(iii) \(KE = PE \Rightarrow \frac{E}{2} = mgh_0 \Rightarrow h_0 = \frac{E}{2mg} = \frac{100}{2 \times 0.5 \times 10} = \mathbf{10 \text{ m}}\)
(iv) \(\frac{1}{2}mv^2 = E \Rightarrow v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = \mathbf{20 \text{ m/s}}\)