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 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
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 \(v\):
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
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).
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.
In many physical systems, forces can be classified as conservative or non-conservative.
When only conservative forces do work, the total mechanical energy (sum of kinetic and potential energies) of the system remains constant:
This principle allows us to analyze motion by equating energies at different points, avoiding direct force calculations.
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.
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.
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.
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.
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.
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.
When to use: When deciding whether to use conservation of mechanical energy or work-energy theorem.
When to use: At the start of every problem solving.
When to use: When forces are not constant or acceleration is not uniform.
When to use: In problems involving friction or other dissipative forces.
When to use: Before starting calculations to visualize the problem.
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