Physics by Lamhi: Not Your Boring Physics

CLASS 11 · CHAPTER 5 · MECHANICS

QUICK REVISION

Work, Energy and Power

Every key result from this chapter, boxed and ready for a last look before the exam. No derivations here, just what to recall and when to use it — for the full explanation, see the detailed notes.

1. Work Done by a Constant Force

W=F⋅d=Fdcos⁡θW = \mathbf{F}\cdot\mathbf{d} = Fd\cos\theta
  1. θ < 90°: force has a component along the motion, work is positive (gravity on a falling stone).
  2. θ = 180°: force opposes the motion, work is negative (friction on a sliding block, gravity on a rising ball).
  3. θ = 90°: force is perpendicular to the motion, work is zero (tension in a conical pendulum, normal force on level motion).

2. Work by a Variable Force & a Spring

Area under the F–x graph:

W=∫xixfF(x) dxW = \int_{x_i}^{x_f} F(x)\,dx

Hooke’s law: F=−kxF=-kx. Work to stretch a spring by xx from natural length:

Work to stretch a spring
W=12kx2W = \dfrac{1}{2}kx^2

3. Kinetic Energy & the Work-Energy Theorem

Kinetic energy: K=12mv2K=\tfrac12 mv^2.

Work-Energy Theorem
Wnet=Kf−Ki=ΔKW_{net} = K_f - K_i = \Delta K

Net work by all forces equals the change in KE — holds for constant or variable force.

4. Potential Energy & Conservative Forces

Gravitational PE
U(h)=mghU(h) = mgh
  1. Conservative force (e.g. gravity): work between two points is path-independent; work around a closed loop is zero.
  2. Non-conservative / dissipative (e.g. friction): work depends on path length, lost as heat, cannot be recovered.

5. Spring PE & Conservation of Mechanical Energy

Spring PE
U(x)=12kx2U(x) = \dfrac{1}{2}kx^2
ΔK=−ΔU    ⟹    K+U=constant\Delta K = -\Delta U \;\;\Longrightarrow\;\; K + U = \text{constant}
  1. No friction (only conservative forces act): mechanical energy K + U stays constant.
  2. With friction (rough incline, angle θ, coefficient μ): slides only if tan⁡θ>μ\tan\theta > \mu; mechanical energy visibly shrinks as it converts to heat, but mechanical + heat together stays constant.
K(s)=U0−U(s)−Q(s)K(s) = U_0 - U(s) - Q(s)

6. Power

Pavg=WtP_{avg} = \dfrac{W}{t}
Instantaneous Power
P=dWdt=F⋅vP = \dfrac{dW}{dt} = \mathbf{F}\cdot\mathbf{v}

7. Elastic Collisions (1D)

Momentum conserved (every collision):

m1u1+m2u2=m1v1+m2v2m_1u_1+m_2u_2 = m_1v_1+m_2v_2
v1=(m1−m2)u1+2m2u2m1+m2v2=(m2−m1)u2+2m1u1m1+m2v_1 = \dfrac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2} \qquad v_2 = \dfrac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2}
  1. Relative velocity of separation = relative velocity of approach: u1+v1=u2+v2u_1+v_1=u_2+v_2.
  2. Equal masses (m₁ = m₂): velocities simply exchange, v1=u2, v2=u1v_1=u_2,\ v_2=u_1.

8. Perfectly Inelastic Collisions

Bodies stick together, move off with one common velocity v — maximum KE loss momentum conservation allows:

m1u1+m2u2=(m1+m2)vm_1u_1+m_2u_2=(m_1+m_2)v

Heat produced (KE lost):

Q=Ki−KfQ = K_i-K_f

Full derivations and worked examples: detailed notes →