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Work, Energy and Power - Kinetic and Potential Energy

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Kinetic Energy (KK) is the energy possessed by an object due to its motion. For a body of mass mm moving with velocity vv, it is calculated as the work required to accelerate the body from rest.

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Potential Energy (UU) is the energy stored in a system by virtue of the relative positions of its parts or its configuration. In a conservative field, work done is stored as potential energy: W=−ΔUW = -\Delta U.

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Gravitational Potential Energy (UgU_g) near the Earth's surface is given by mghmgh, where hh is the height above a chosen reference level (datum).

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Elastic Potential Energy (UsU_s) is the energy stored in a deformed elastic object (like a spring), given by the work done against the restoring force F=−kxF = -kx.

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The Work-Energy Theorem states that the net work done by all forces (conservative and non-conservative) on a particle is equal to the change in its kinetic energy: Wnet=ΔKW_{net} = \Delta K.

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The Law of Conservation of Mechanical Energy states that if only conservative forces do work, the total mechanical energy (E=K+UE = K + U) of the system remains constant.

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Conservative forces (e.g., gravitational, electrostatic) are path-independent and work done over a closed loop is zero. Non-conservative forces (e.g., friction, air resistance) are path-dependent and dissipate energy as heat.

📐Formulae

K=12mv2K = \frac{1}{2}mv^2

K=p22m (where p=mv is linear momentum)K = \frac{p^2}{2m} \text{ (where } p = mv \text{ is linear momentum)}

Ug=mghU_g = mgh

Us=12kx2 (where k is the spring constant)U_s = \frac{1}{2}kx^2 \text{ (where } k \text{ is the spring constant)}

Wnet=Kf−Ki=12m(v2−u2)W_{net} = K_f - K_i = \frac{1}{2}m(v^2 - u^2)

Etotal=K+U=constant (if Wnc=0)E_{total} = K + U = \text{constant (if } W_{nc} = 0\text{)}

F=−dUdx (Relationship between conservative force and potential energy)F = -\frac{dU}{dx} \text{ (Relationship between conservative force and potential energy)}

💡Examples

Problem 1:

A block of mass 2 kg2 \text{ kg} is pushed against a horizontal spring of force constant k=500 N/mk = 500 \text{ N/m}, compressing it by 10 cm10 \text{ cm}. When released, the block moves on a frictionless surface. Calculate the velocity of the block when it leaves the spring.

Solution:

  1. Initial Potential Energy of the spring: Ui=12kx2=12(500)(0.1)2=2.5 JU_i = \frac{1}{2}kx^2 = \frac{1}{2}(500)(0.1)^2 = 2.5 \text{ J}.
  2. Initial Kinetic Energy: Ki=0K_i = 0.
  3. By Conservation of Mechanical Energy: Ui+Ki=Uf+KfU_i + K_i = U_f + K_f.
  4. When the block leaves the spring, Uf=0U_f = 0. So, 2.5+0=0+12mv22.5 + 0 = 0 + \frac{1}{2}mv^2.
  5. 2.5=12(2)v2  ⟹  v2=2.5  ⟹  v=2.5≈1.58 m/s2.5 = \frac{1}{2}(2)v^2 \implies v^2 = 2.5 \implies v = \sqrt{2.5} \approx 1.58 \text{ m/s}.

Explanation:

The elastic potential energy stored in the compressed spring is completely converted into the kinetic energy of the block as the spring returns to its natural length.

Problem 2:

The momentum of a body is increased by 20%20\%. Calculate the percentage increase in its kinetic energy.

Solution:

  1. Let initial momentum be p1p_1 and initial kinetic energy be K1=p122mK_1 = \frac{p_1^2}{2m}.
  2. New momentum p2=p1+20% of p1=1.2p1p_2 = p_1 + 20\% \text{ of } p_1 = 1.2p_1.
  3. New kinetic energy K2=p222m=(1.2p1)22m=1.44p122m=1.44K1K_2 = \frac{p_2^2}{2m} = \frac{(1.2p_1)^2}{2m} = 1.44 \frac{p_1^2}{2m} = 1.44K_1.
  4. Percentage increase =K2−K1K1×100=1.44K1−K1K1×100=44%= \frac{K_2 - K_1}{K_1} \times 100 = \frac{1.44K_1 - K_1}{K_1} \times 100 = 44\%.

Explanation:

Since kinetic energy is proportional to the square of the momentum (K∝p2K \propto p^2), a linear increase in momentum results in a quadratic increase in kinetic energy.

Problem 3:

A ball of mass 0.5 kg0.5 \text{ kg} is dropped from a height of 10 m10 \text{ m}. Find its kinetic energy and velocity just before hitting the ground. (Take g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

  1. At height hh, Potential Energy U=mgh=0.5×9.8×10=49 JU = mgh = 0.5 \times 9.8 \times 10 = 49 \text{ J}.
  2. By Law of Conservation of Energy, Utop=KbottomU_{top} = K_{bottom}.
  3. K=49 JK = 49 \text{ J}.
  4. Since K=12mv2K = \frac{1}{2}mv^2, then 49=12(0.5)v2  ⟹  v2=490.25=196  ⟹  v=196=14 m/s49 = \frac{1}{2}(0.5)v^2 \implies v^2 = \frac{49}{0.25} = 196 \implies v = \sqrt{196} = 14 \text{ m/s}.

Explanation:

As the ball falls, its gravitational potential energy is converted into kinetic energy. At the point of impact, all initial potential energy (relative to the ground) has become kinetic energy.

Kinetic and Potential Energy - Revision Notes & Key Formulas | ICSE Class 11 Physics