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Work energy theorem and conservation

Introduction

In the study of mechanics, two fundamental concepts-work and energy-play a central role in understanding how forces cause motion and how motion changes. Work quantifies the effect of a force acting over a distance, while energy measures the capacity of a system to perform work. The work-energy theorem connects these two ideas by stating that the net work done on an object results in a change in its kinetic energy.

Moreover, the conservation of mechanical energy principle allows us to analyze systems where energy transforms between kinetic and potential forms without loss, simplifying complex problems. These concepts are not only foundational in physics but also powerful tools in solving competitive exam problems efficiently.

This section will build your understanding from the ground up, starting with the definitions of work and energy, moving through the work-energy theorem, and culminating in the conservation of mechanical energy. Along the way, you will encounter clear explanations, diagrams, and worked examples to solidify your grasp.

Work and Energy

Work is done when a force causes an object to move. Specifically, work is defined as the product of the component of the force in the direction of displacement and the magnitude of that displacement.

Mathematically, if a constant force \(\vec{F}\) acts on an object causing a displacement \(\vec{d}\), the work done \(W\) is

Work Done by a Constant Force

\[W = \vec{F} \cdot \vec{d} = F d \cos\theta\]

Work is the dot product of force and displacement vectors

W = work (Joules)
F = magnitude of force (Newtons)
d = displacement (meters)
\(\theta\) = angle between force and displacement

Here, \(\theta\) is the angle between the force and displacement vectors. If the force is in the same direction as displacement, \(\theta = 0^\circ\) and \(\cos 0^\circ = 1\), so work is positive. If the force opposes displacement, \(\theta = 180^\circ\) and work is negative, indicating energy is taken away from the system.

Kinetic energy is the energy an object possesses due to its motion. It depends on the mass \(m\) of the object and the square of its velocity \(v\):

Kinetic Energy

\[K = \frac{1}{2} m v^2\]

Energy due to motion of an object

K = kinetic energy (Joules)
m = mass (kg)
v = velocity (m/s)

Potential energy is energy stored due to an object's position or configuration. The most common form is gravitational potential energy near Earth's surface, given by

Gravitational Potential Energy

U = m g h

Energy stored due to height in a gravitational field

U = potential energy (Joules)
m = mass (kg)
g = acceleration due to gravity (9.8 m/s²)
h = height above reference point (meters)

Both kinetic and potential energies are scalars, meaning they have magnitude but no direction. The unit of energy and work in the metric system is the joule (J), where 1 joule equals 1 newton-meter (N·m).

Block Inclined plane Force (gravity component) Displacement Potential Energy (U) decreases as block slides down Kinetic Energy (K) increases

Work-Energy Theorem

The work-energy theorem states that the net work done by all forces acting on a particle equals the change in its kinetic energy. This theorem provides a powerful link between force, displacement, and motion without directly solving Newton's equations.

graph TD    A[Start: Newton's Second Law: F = m a] --> B[Express acceleration as dv/dt]    B --> C[Multiply both sides by velocity v]    C --> D[Rewrite as d(1/2 m v²)/dt = F · v]    D --> E[Integrate over time to get work done]    E --> F[Net work done = Change in kinetic energy]

Derivation Outline:

Consider a particle of mass \(m\) moving under a net force \(\vec{F}\). Newton's second law gives

\(\vec{F} = m \vec{a}\), where \(\vec{a}\) is acceleration.

Acceleration can be written as the derivative of velocity \(\vec{v}\) with respect to time:

\(\vec{a} = \frac{d\vec{v}}{dt}\).

Dot-multiplied by velocity \(\vec{v}\), we get

\(\vec{F} \cdot \vec{v} = m \frac{d\vec{v}}{dt} \cdot \vec{v} = m \frac{d}{dt} \left(\frac{v^2}{2}\right)\).

Since \(\vec{v} \cdot \vec{v} = v^2\), the left side represents the instantaneous power (rate of doing work).

Integrating over the time interval from \(t_i\) to \(t_f\), or equivalently over displacement,

\[W_{net} = \int \vec{F} \cdot d\vec{r} = \int m \frac{d}{dt} \left(\frac{v^2}{2}\right) dt = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 = \Delta K\]

This shows that the net work done on the particle equals the change in its kinetic energy.

The work-energy theorem is especially useful when forces vary or when acceleration is not constant, allowing us to bypass complicated force and acceleration calculations.

Conservation of Mechanical Energy

In many physical systems, forces can be classified as conservative or non-conservative.

  • Conservative forces (like gravity and spring force) have the property that the work done by them depends only on the initial and final positions, not on the path taken. They store energy as potential energy.
  • Non-conservative forces (like friction and air resistance) dissipate mechanical energy, usually converting it into heat or sound.

When only conservative forces do work, the total mechanical energy (sum of kinetic and potential energies) of the system remains constant:

Conservation of Mechanical Energy

\[K_i + U_i = K_f + U_f\]

Total mechanical energy remains constant if only conservative forces act

K = kinetic energy (J)
U = potential energy (J)
i = initial state
f = final state

This principle allows us to analyze motion by equating energies at different points, avoiding direct force calculations.

Highest Point Lowest Point Potential Energy (max) Kinetic Energy (max)

In the pendulum shown, at the highest point, the bob has maximum potential energy and zero kinetic energy (momentarily at rest). At the lowest point, potential energy is minimum, and kinetic energy is maximum. The total mechanical energy remains constant if air resistance and friction at the pivot are negligible.

When non-conservative forces like friction act, mechanical energy decreases, and the conservation principle no longer holds exactly. Instead, the work-energy theorem including non-conservative work must be used.

Worked Example 1: Using Work-Energy Theorem to Find Velocity

Example 1: Velocity of Block Sliding Down an Incline Easy
A block of mass 2 kg slides down a frictionless incline of height 5 m. Find the speed of the block at the bottom of the incline using the work-energy theorem.

Step 1: Identify forces and energies. Gravity does work causing the block to accelerate down.

Step 2: Initial velocity \(v_i = 0\) (starting from rest), initial kinetic energy \(K_i = 0\).

Step 3: Calculate initial potential energy \(U_i = m g h = 2 \times 9.8 \times 5 = 98\, \text{J}\).

Step 4: At the bottom, height \(h_f = 0\), so potential energy \(U_f = 0\).

Step 5: Using conservation of mechanical energy (no friction):

\(K_i + U_i = K_f + U_f\)

\(0 + 98 = \frac{1}{2} \times 2 \times v_f^2 + 0\)

Step 6: Solve for \(v_f\):

\(98 = v_f^2\)

\(v_f = \sqrt{98} = 9.9\, \text{m/s}\)

Answer: The block's speed at the bottom is approximately 9.9 m/s.

Block (2 kg) Inclined plane (height 5 m) Gravity component Displacement

Worked Example 2: Calculating Work Done by a Variable Force

Example 2: Work Done by a Variable Force Medium
A force \(F(x) = 3x^2\) N acts on a particle moving along the x-axis from \(x = 0\) m to \(x = 4\) m. Calculate the work done by this force.

Step 1: Recognize that force varies with position, so work is the integral of force over displacement:

\[ W = \int_{x_i}^{x_f} F(x) \, dx \]

Step 2: Substitute \(F(x) = 3x^2\), \(x_i = 0\), \(x_f = 4\):

\[ W = \int_0^4 3x^2 \, dx = 3 \int_0^4 x^2 \, dx \]

Step 3: Integrate:

\[ 3 \left[ \frac{x^3}{3} \right]_0^4 = \left[ x^3 \right]_0^4 = 4^3 - 0 = 64\, \text{J} \]

Answer: The work done by the force is 64 joules.

Worked Example 3: Energy Conservation with Friction

Example 3: Energy Conservation with Friction Hard
A 5 kg block slides down a 10 m long incline at 30° to the horizontal. The coefficient of kinetic friction between the block and incline is 0.1. Find the speed of the block at the bottom of the incline.

Step 1: Calculate vertical height \(h\):

\(h = L \sin \theta = 10 \times \sin 30^\circ = 10 \times 0.5 = 5\, \text{m}\)

Step 2: Calculate gravitational potential energy lost:

\(U_i - U_f = m g h = 5 \times 9.8 \times 5 = 245\, \text{J}\)

Step 3: Calculate friction force \(f_k\):

Normal force \(N = m g \cos \theta = 5 \times 9.8 \times \cos 30^\circ = 5 \times 9.8 \times 0.866 = 42.4\, \text{N}\)

Friction force \(f_k = \mu_k N = 0.1 \times 42.4 = 4.24\, \text{N}\)

Step 4: Work done by friction (negative):

\(W_f = - f_k \times d = -4.24 \times 10 = -42.4\, \text{J}\)

Step 5: Net work done on block equals change in kinetic energy:

\(W_{net} = \Delta K = (U_i - U_f) + W_f = 245 - 42.4 = 202.6\, \text{J}\)

Step 6: Calculate final velocity \(v_f\):

\[ \Delta K = \frac{1}{2} m v_f^2 - 0 = 202.6 \]

\[ v_f = \sqrt{\frac{2 \times 202.6}{5}} = \sqrt{81.04} = 9.0\, \text{m/s} \]

Answer: The block's speed at the bottom is approximately 9.0 m/s.

Worked Example 4: Potential Energy Change in a Spring System

Example 4: Spring Potential Energy and Kinetic Energy Medium
A block of mass 0.5 kg is compressed against a spring with spring constant \(k = 200\, \text{N/m}\) by 0.1 m and then released on a frictionless surface. Find the speed of the block when the spring returns to its natural length.

Step 1: Calculate initial potential energy stored in the spring:

\[ U = \frac{1}{2} k x^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 1\, \text{J} \]

Step 2: At natural length, spring potential energy is zero; all energy converts to kinetic energy:

\[ K = \frac{1}{2} m v^2 = 1\, \text{J} \]

Step 3: Solve for velocity \(v\):

\[ v = \sqrt{\frac{2 \times 1}{0.5}} = \sqrt{4} = 2\, \text{m/s} \]

Answer: The block's speed when the spring is uncompressed is 2 m/s.

Worked Example 5: Work Done by Gravity in Orbital Motion

Example 5: Work Done by Gravity in Orbital Motion Hard
A satellite of mass 1000 kg moves from a circular orbit of radius \(7 \times 10^6\) m to a higher circular orbit of radius \(1.4 \times 10^7\) m around Earth. Calculate the work done by Earth's gravity during this process.

Step 1: Gravitational force is conservative; work done equals change in gravitational potential energy.

Gravitational potential energy at radius \(r\) is

\[ U = - \frac{G M m}{r} \]

where \(G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\), \(M = 5.98 \times 10^{24} \, \text{kg}\) (Earth's mass), \(m = 1000\, \text{kg}\).

Step 2: Calculate initial and final potential energies:

\[ U_i = - \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times 1000}{7 \times 10^6} = -5.7 \times 10^{10} \, \text{J} \]

\[ U_f = - \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times 1000}{1.4 \times 10^7} = -2.85 \times 10^{10} \, \text{J} \]

Step 3: Work done by gravity is change in potential energy:

\[ W = U_i - U_f = (-5.7 \times 10^{10}) - (-2.85 \times 10^{10}) = -2.85 \times 10^{10} \, \text{J} \]

Negative work indicates gravity does work against the satellite's motion as it moves to a higher orbit.

Answer: The work done by gravity is \(-2.85 \times 10^{10}\) joules.

Formula Bank

Work Done by a Constant Force
\[ W = \vec{F} \cdot \vec{d} = F d \cos\theta \]
where: \(W\) = work (J), \(F\) = force (N), \(d\) = displacement (m), \(\theta\) = angle between force and displacement
Kinetic Energy
\[ K = \frac{1}{2} m v^2 \]
where: \(K\) = kinetic energy (J), \(m\) = mass (kg), \(v\) = velocity (m/s)
Potential Energy (Gravitational)
\[ U = m g h \]
where: \(U\) = potential energy (J), \(m\) = mass (kg), \(g\) = acceleration due to gravity (9.8 m/s²), \(h\) = height (m)
Work-Energy Theorem
\[ W_{net} = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]
where: \(W_{net}\) = net work (J), \(m\) = mass (kg), \(v_i\) = initial velocity (m/s), \(v_f\) = final velocity (m/s)
Conservation of Mechanical Energy
\[ K_i + U_i = K_f + U_f \]
where: \(K\) = kinetic energy (J), \(U\) = potential energy (J), subscripts \(i\) and \(f\) denote initial and final states
Work Done by Variable Force
\[ W = \int_{x_i}^{x_f} F(x) \, dx \]
where: \(W\) = work (J), \(F(x)\) = force as function of position, \(x_i\), \(x_f\) = initial and final positions (m)

Tips & Tricks

Tip: Always identify if forces are conservative or non-conservative before applying energy conservation.

When to use: When deciding whether to use conservation of mechanical energy or work-energy theorem.

Tip: Convert all units to SI (metric) before calculations to avoid errors.

When to use: At the start of every problem solving.

Tip: Use energy methods instead of force methods for problems involving variable forces or complicated motion.

When to use: When forces are not constant or acceleration is not uniform.

Tip: Remember that work done by friction is negative and reduces mechanical energy.

When to use: In problems involving friction or other dissipative forces.

Tip: Sketch the problem scenario and mark forces, displacement, and energy states clearly.

When to use: Before starting calculations to visualize the problem.

Common Mistakes to Avoid

❌ Confusing work done by conservative forces with total work done.
✓ Remember that work done by conservative forces changes potential energy, while total work includes all forces.
Why: Students often overlook non-conservative forces like friction.
❌ Using conservation of mechanical energy when friction or air resistance is present.
✓ Use work-energy theorem including work done by non-conservative forces instead.
Why: Misunderstanding the conditions for energy conservation.
❌ Ignoring the angle between force and displacement when calculating work.
✓ Always include the cosine of the angle between force and displacement vectors.
Why: Forces not aligned with displacement do less work.
❌ Mixing units, such as using cm instead of meters or km/h instead of m/s.
✓ Convert all units to SI units before calculations.
Why: Leads to incorrect numerical answers.
❌ Assuming kinetic energy is zero at the highest point without checking velocity.
✓ Calculate velocity explicitly; kinetic energy depends on velocity squared.
Why: Velocity may not be zero if external forces act.
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