Work
In physics, work is done when a force causes displacement. W = vecF . vecs = Fs cos theta
Special Cases
theta = 0:W = Fs(maximum positive work)theta = 90:W = 0(force perpendicular to displacement)theta = 180:W = -Fs(maximum negative work)
Work Done by Variable Force
W = int_(x_1)^(x_2) F(x) dx
This equals the area under the F-x curve.
Units of Work
SI: joule (J). 1 J = 1 N m.
CGS: erg. 1 J = 10^7 erg.
Kinetic Energy
Energy possessed by a body due to its motion.
K = (1/2) mv^2
Derivation
K = int F dx = int m (dv/dt) dx = m int v dv = (1/2) mv^2
Potential Energy
Energy possessed by a body due to its position or configuration.
Gravitational Potential Energy
U = mgh (near Earth's surface)
Elastic Potential Energy
U = (1/2) k x^2 (for a spring, where k is spring constant)
Work-Energy Theorem
The net work done on a body equals the change in its kinetic energy.
W_net = Delta K = K_f - K_i = (1/2) m(v^2 - u^2)
Proof: W = Fs = ma * s = m * (v^2-u^2)/2 = (1/2)m(v^2-u^2) = Delta K
Conservative and Non-Conservative Forces
| Conservative | Non-Conservative |
|---|---|
| Work done is path independent | Work done depends on path |
| Work done in closed loop is zero | Work done in closed loop is not zero |
| Potential energy defined | No potential energy |
| Examples: gravity, spring force, electrostatic | Examples: friction, air resistance, viscous force |
Law of Conservation of Energy
Energy can neither be created nor destroyed; it can only transform from one form to another. The total energy of an isolated system remains constant.
Example (Freely falling body): (1/2) mv^2 + mgh = constant
At height h: PE = mgh, KE = 0, TE = mgh.
At ground: PE = 0, KE = (1/2) mv^2 = mgh, TE = mgh.
Power
Rate of doing work.
P = W/t = vecF . vecv (for constant force)
Units
SI: watt (W). 1 W = 1 J/s.
1 hp = 746 W.
1 kWh = 3.6 x 10^6 J.
Elastic and Inelastic Collisions
Elastic Collision
Both momentum and kinetic energy are conserved.
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
(1/2) m_1 u_1^2 + (1/2) m_2 u_2^2 = (1/2) m_1 v_1^2 + (1/2) m_2 v_2^2
Velocities after collision:
v_1 = ((m_1 - m_2)/(m_1 + m_2)) u_1 + (2m_2/(m_1 + m_2)) u_2
v_2 = (2m_1/(m_1 + m_2)) u_1 - ((m_1 - m_2)/(m_1 + m_2)) u_2
Special cases:
- Equal masses: Velocities are exchanged.
v_1 = u_2,v_2 = u_1. - Target at rest (
u_2 = 0):v_1 = ((m_1 - m_2)/(m_1 + m_2)) u_1,v_2 = (2m_1/(m_1 + m_2)) u_1.
Inelastic Collision
Only momentum is conserved. Kinetic energy is not conserved (lost as heat, sound, deformation).
m_1 u_1 + m_2 u_2 = (m_1 + m_2)v (for perfectly inelastic where bodies stick together)
Coefficient of restitution: e = (v_2 - v_1)/(u_1 - u_2). For elastic e = 1, inelastic 0 < e < 1, perfectly inelastic e = 0.
Worked Examples
Example 1: A 10 kg block is pulled 5 m on a rough surface with mu_k = 0.2 by a 50 N force at 37 degrees. Find work done by each force and net work.
Solution: W_F = 50*5*cos37 = 200 J. W_f = f_k * s * cos180 = -mu_k mg s = -0.2*10*10*5 = -100 J. W_N = W_mg = 0. Net work = 200 - 100 = 100 J.
Example 2: A spring of k = 500 N/m is compressed by 0.1 m. Find elastic PE.
Solution: PE = (1/2)kx^2 = 0.5*500*0.01 = 2.5 J.
Example 3: A 1 kg ball moving at 4 m/s collides elastically with a stationary 2 kg ball. Find velocities after collision.
Solution: v_1 = (1-2)/(1+2) * 4 = -4/3 m/s. v_2 = (2*1)/(1+2) * 4 = 8/3 m/s.
Common Mistakes
- Work and energy are scalars: No direction, despite being derived from vectors.
- Negative work: Force opposite to displacement does negative work.
- Conservation of ME: Only valid when only conservative forces act.
- Coefficient of restitution:
e = (v_2 - v_1)/(u_1 - u_2)— note the order carefully.
ISC Exam Focus
- Theory (70%): Work-energy theorem derivation, conservative forces, conservation of energy, collision types.
- Application (30%): Numerical problems involving work, energy, power, and collisions.
- ISC typical questions: "A block slides down an incline ... find velocity/speed using energy conservation."
- 4-6 mark collision numericals are common.
Self-Test Questions
Q1: Define work. What is the work done if the force is perpendicular to displacement? Answer: Work = force x displacement in force direction. If perpendicular, work = 0.
Q2: A 2 kg body falls from a height of 10 m. Find its KE just before hitting ground.
Answer: By conservation of energy: KE = PE = mgh = 2*10*10 = 200 J.
Q3: A 60 W bulb is used for 5 hours. Find energy consumed in kWh.
Answer: Energy = P*t = 60*5 = 300 Wh = 0.3 kWh.
Q4: State the work-energy theorem.
Answer: Net work done equals change in kinetic energy: W = (1/2)mv^2 - (1/2)mu^2.
Q5: In a perfectly inelastic collision, two bodies stick and move together after collision. A 2 kg body moving at 4 m/s hits a stationary 2 kg body. Find the common velocity.
Answer: m_1 u_1 + m_2 u_2 = (m_1+m_2)v. 2*4 + 0 = 4v => v = 2 m/s.
Q6: Differentiate between conservative and non-conservative forces. Answer: Conservative: path independent, closed loop work zero, PE defined. Non-conservative: path dependent, closed loop work non-zero, no PE.
