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. 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.
• Formula: $U = mgh$
• Explanation: Work done against gravity in lifting a mass $m$ to a height $h$.
• Formula: $U = \frac{1}{2} k x^2$
• Explanation: Energy stored in a deformed elastic body (like a spring stretched by distance $x$).
• 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:
$$F_s = -k x$$
Figure 1: Spring-mass system at rest ($x=0$) and stretched by displacement $x$.
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 $K = \frac{1}{2} m v^2$.
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
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.
$$W_{\text{net}} = \Delta K = K_f – K_i = \frac{1}{2}m v^2 – \frac{1}{2}m u^2$$
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
