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WORK POWER AND ENERGY – II

Work, Power, and Energy: Energy Concepts, Spring Mechanics & Work-Energy Theorem

Welcome to the comprehensive guide on Energy, Spring Potential Energy, Kinetic Energy Derivations, and the Work-Energy Theorem. This study module covers fundamental derivations, mathematical foundations, visual diagrams, conceptual questions, and step-by-step solved numerical problems.

Table of Contents

  1. 1. Definition of Energy
  2. 2. Potential Energy & Case Analysis
  3. 3. Elastic Potential Energy of a Spring
  4. 4. Kinetic Energy & Derivation
  5. 5. Work-Energy Theorem
  6. 6. Conceptual Questions & Solved Numerical Examples

1. Definition of Energy

Definition: Energy is defined as the total capacity of a physical system to perform work. It is a scalar quantity, meaning it possesses magnitude but no direction.

SI Unit: Joule ($\text{J}$), where $1\text{ J} = 1\text{ N}\cdot\text{m} = 1\text{ kg}\cdot\text{m}^2/\text{s}^2$

CGS Unit: Erg, where $1\text{ Joule} = 10^7\text{ Ergs}$

Dimensional Formula: $[\text{M}^1 \text{L}^2 \text{T}^{-2}]$

2. Potential Energy & Case Analysis

Potential Energy ($U$) is the energy stored within a body or system by virtue of its position, configuration, or state of strain against a conservative force.

Case 1: Gravitational Potential Energy

Formula: $U = mgh$

Explanation: Work done against gravity in lifting a mass $m$ to a height $h$.

Case 2: Elastic Potential Energy

Formula: $U = \frac{1}{2} k x^2$

Explanation: Energy stored in a deformed elastic body (like a spring stretched by distance $x$).

Case 3: Electrostatic Potential Energy

Formula: $U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}$

Explanation: Energy stored due to the relative configuration of two point charges separated by distance $r$.

3. Elastic Potential Energy of a Spring

According to Hooke’s Law, when an ideal spring is stretched or compressed by a displacement $x$ from its mean position, the restoring force $F_s$ is given by:

$$F_s = -k x$$

m Equilibrium ($x=0$) m Displacement ($x$)

Figure 1: Spring-mass system at rest ($x=0$) and stretched by displacement $x$.

Mathematical Derivation:

1. The small work done $dW$ by an external force to stretch the spring by an infinitesimal distance $dx$ against the restoring force is:

$$dW = F_{\text{ext}} \, dx = kx \, dx$$

2. Integrating from initial position $x = 0$ to final elongation $x$:

$$W = \int_{0}^{x} kx \, dx = k \left[ \frac{x^2}{2} \right]_{0}^{x} = \frac{1}{2} k x^2$$

3. Stored Elastic Potential Energy ($U$): $U = \frac{1}{2} k x^2$

4. Kinetic Energy & Derivation

Kinetic Energy ($K$) is the energy possessed by an object due to its motion. If a body of mass $m$ is moving with velocity $v$, its kinetic energy is given by $K = \frac{1}{2} m v^2$.

Step-by-Step Derivation:

1. Consider a body of mass $m$ initially at rest ($u = 0$). A constant force $F$ acts on it, accelerating it to velocity $v$ over displacement $s$.

2. Using Newton’s second law of motion: $F = m a$

3. From the third equation of motion ($v^2 = u^2 + 2as$):

$$v^2 = 0 + 2as \implies a s = \frac{v^2}{2}$$

4. Substituting force and displacement into the Work formula ($W = F s$):

$$W = (m a) s = m (a s) = m \left( \frac{v^2}{2} \right) = \frac{1}{2} m v^2$$

5. Since work done equals kinetic energy gained: $K = \frac{1}{2} m v^2$

5. Work-Energy Theorem

Statement:

The net work done by all forces (conservative, non-conservative, internal, and external) acting on a body is equal to the net change in its kinetic energy.

$$W_{\text{net}} = \Delta K = K_f – K_i = \frac{1}{2}m v^2 – \frac{1}{2}m u^2$$

Derivation for Variable Force:

1. Instantaneous rate of change of kinetic energy:

$$\frac{dK}{dt} = \frac{d}{dt} \left(\frac{1}{2} m v^2\right) = m v \frac{dv}{dt} = m a v = F v = F \frac{dx}{dt}$$

2. Canceling $dt$ on both sides yields: $dK = F dx$

3. Integrating from initial state $(x_i, K_i)$ to final state $(x_f, K_f)$:

$$\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F dx \implies K_f – K_i = W_{\text{net}}$$


6. Conceptual Questions & Solved Numerical Examples

Part A: Conceptual Understanding Questions

Q1. Can a body have kinetic energy without having momentum?

No. Kinetic energy is related to momentum by $K = \frac{p^2}{2m}$. If momentum $p = 0$, velocity $v = 0$, so kinetic energy must be zero.

Q2. If the momentum of a body is doubled, by what factor does its kinetic energy increase?

Since $K \propto p^2$, doubling the momentum ($2p$) increases kinetic energy by $2^2 = \mathbf{4}$ times (quadrupled).

Q3. Can potential energy of a body be negative? Give an example.

Yes. Potential energy depends on the reference frame. For instance, gravitational potential energy below a chosen zero-level surface is negative ($U = -mgh$).

Q4. Does the Work-Energy Theorem hold true in non-inertial frames of reference?

Yes, provided the work done by pseudo forces is included in calculating $W_{\text{net}}$.

Q5. What happens to the potential energy of a spring when it is compressed vs. stretched?

In both cases, potential energy increases because $U = \frac{1}{2} k x^2 \ge 0$, regardless of whether $x$ is positive or negative.

Q6. A heavy body and a light body have equal kinetic energies. Which one has greater momentum?

Since $p = \sqrt{2mK}$, for equal kinetic energy $K$, momentum is proportional to $\sqrt{m}$. The heavier body has greater momentum.

Q7. Can kinetic energy ever be negative?

No. Mass $m > 0$ and $v^2 \ge 0$, so kinetic energy is strictly non-negative ($K \ge 0$).

Q8. When a bullet gets embedded in a wooden block, what happens to its kinetic energy?

Most kinetic energy is converted into heat, sound, and non-conservative work doing structural deformation.

Q9. How does spring stiffness constant ($k$) affect work required to stretch a spring?

A stiffer spring (larger $k$) requires proportionally more work ($W = \frac{1}{2} k x^2$) for the same displacement $x$.

Q10. Is kinetic energy dependent on the frame of reference?

Yes. Velocity is relative to the frame of reference, so kinetic energy value varies by observer motion.

Part B: Solved Numerical Examples

Problem 1: Kinetic Energy Calculation

Question: Calculate the kinetic energy of an object of mass $10\text{ kg}$ moving with a uniform velocity of $6\text{ m/s}$.

Given Data: Mass $m = 10\text{ kg}$, Velocity $v = 6\text{ m/s}$

Formula: $K = \frac{1}{2} m v^2$

Step-by-Step Solution:

$$K = \frac{1}{2} \times 10\text{ kg} \times (6\text{ m/s})^2$$

$$K = 5 \times 36 = 180\text{ J}$$

Final Answer: $180\text{ Joules}$

Problem 2: Elastic Potential Energy of a Compressed Spring

Question: A spring with a spring constant $k = 500\text{ N/m}$ is compressed by $0.05\text{ m}$ from its natural length. Find the elastic potential energy stored in the spring.

Given Data: Spring constant $k = 500\text{ N/m}$, Compression $x = 0.05\text{ m}$

Formula: $U = \frac{1}{2} k x^2$

Step-by-Step Solution:

$$U = \frac{1}{2} \times 500 \times (0.05)^2$$

$$U = 250 \times 0.0025 = 0.625\text{ J}$$

Final Answer: $0.625\text{ Joules}$

Problem 3: Work-Energy Theorem for a Braking Vehicle

Question: A car of mass $1000\text{ kg}$ traveling at a speed of $20\text{ m/s}$ is brought to rest by applying brakes. Calculate the net work done on the car by the braking force.

Given Data: Mass $m = 1000\text{ kg}$, Initial speed $u = 20\text{ m/s}$, Final speed $v = 0\text{ m/s}$

Formula (Work-Energy Theorem): $W_{\text{net}} = \Delta K = \frac{1}{2} m v^2 – \frac{1}{2} m u^2$

Step-by-Step Solution:

$$W_{\text{net}} = 0 – \frac{1}{2} \times 1000 \times (20)^2$$

$$W_{\text{net}} = -500 \times 400 = -200,000\text{ J} = -200\text{ kJ}$$

Final Answer: $-200\text{ kJ}$ (negative sign denotes energy dissipated by friction)

Problem 4: Relation Between Momentum & Kinetic Energy

Question: A body of mass $4\text{ kg}$ has a linear momentum of $20\text{ kg}\cdot\text{m/s}$. Find its kinetic energy.

Given Data: Mass $m = 4\text{ kg}$, Momentum $p = 20\text{ kg}\cdot\text{m/s}$

Formula: $K = \frac{p^2}{2m}$

Step-by-Step Solution:

$$K = \frac{(20)^2}{2 \times 4} = \frac{400}{8} = 50\text{ J}$$

Final Answer: $50\text{ Joules}$

Problem 5: Gravitational Potential Energy

Question: Calculate the potential energy gained by a $5\text{ kg}$ object when raised vertically to a height of $12\text{ m}$. (Take $g = 9.8\text{ m/s}^2$)

Given Data: Mass $m = 5\text{ kg}$, Height $h = 12\text{ m}$, Acceleration due to gravity $g = 9.8\text{ m/s}^2$

Formula: $U = mgh$

Step-by-Step Solution:

$$U = 5 \times 9.8 \times 12 = 588\text{ J}$$

Final Answer: $588\text{ Joules}$

Problem 6: Work Required to Change Speed

Question: How much work must be done to accelerate a $2\text{ kg}$ block from $5\text{ m/s}$ to $10\text{ m/s}$ on a frictionless horizontal plane?

Given Data: Mass $m = 2\text{ kg}$, Initial speed $u = 5\text{ m/s}$, Final speed $v = 10\text{ m/s}$

Formula: $W = \Delta K = \frac{1}{2} m (v^2 – u^2)$

Step-by-Step Solution:

$$W = \frac{1}{2} \times 2 \times (10^2 – 5^2)$$

$$W = 1 \times (100 – 25) = 75\text{ J}$$

Final Answer: $75\text{ Joules}$

Problem 7: Spring Extension Work Ratio

Question: It takes $10\text{ J}$ of work to stretch a spring by a displacement $x$. How much total energy is stored in the spring when it is stretched by a total displacement of $3x$ from equilibrium?

Given Data: Initial stored energy $U_1 = 10\text{ J}$ at extension $x$. New extension $x_2 = 3x$.

Formula: $U \propto x^2 \implies \frac{U_2}{U_1} = \left(\frac{x_2}{x_1}\right)^2$

Step-by-Step Solution:

$$\frac{U_2}{10} = \left(\frac{3x}{x}\right)^2 = 3^2 = 9$$

$$U_2 = 9 \times 10 = 90\text{ J}$$

Final Answer: $90\text{ Joules}$

Problem 8: Work Done by Variable Force

Question: A variable force $F(x) = 6x\text{ N}$ acts on a $3\text{ kg}$ mass moving along the x-axis from $x = 0$ to $x = 2\text{ m}$. Find the total change in kinetic energy of the mass.

Given Data: Force function $F(x) = 6x$, Limits $x_i = 0$ to $x_f = 2\text{ m}$

Formula: $\Delta K = W = \int_{x_i}^{x_f} F(x) \, dx$

Step-by-Step Solution:

$$\Delta K = \int_{0}^{2} 6x \, dx = \left[ 3x^2 \right]_{0}^{2}$$

$$\Delta K = 3(2)^2 – 3(0)^2 = 3(4) = 12\text{ J}$$

Final Answer: $12\text{ Joules}$

Problem 9: Velocity from Conservation of Energy

Question: A body of mass $m$ is dropped from rest from a height of $20\text{ m}$. Using energy conservation, calculate its velocity right before striking the ground. (Take $g = 10\text{ m/s}^2$)

Given Data: Height $h = 20\text{ m}$, Initial velocity $u = 0$, $g = 10\text{ m/s}^2$

Formula: Potential Energy Lost = Kinetic Energy Gained $\implies mgh = \frac{1}{2} m v^2$

Step-by-Step Solution:

$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20}$$

$$v = \sqrt{400} = 20\text{ m/s}$$

Final Answer: $20\text{ m/s}$

Problem 10: Percentage Increase in Kinetic Energy

Question: If the velocity of a moving object is increased by $50\%$, determine the percentage increase in its kinetic energy.

Given Data: Initial velocity $v_1 = v$, New velocity $v_2 = v + 0.50v = 1.5v$

Formula: $\text{Percentage Increase} = \left(\frac{K_2 – K_1}{K_1}\right) \times 100\%$

Step-by-Step Solution:

$$K_1 = \frac{1}{2} m v^2$$

$$K_2 = \frac{1}{2} m (1.5v)^2 = 2.25 \left(\frac{1}{2} m v^2\right) = 2.25 K_1$$

$$\text{Percentage Increase} = \frac{2.25 K_1 – K_1}{K_1} \times 100\% = 1.25 \times 100\% = 125\%$$

Final Answer: $125\%$ increase