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In classical mechanics, two fundamental concepts help us understand how forces cause motion and how objects store and transfer energy: work and energy. These ideas are central to analyzing physical systems, from a simple block sliding on a table to complex machines and natural phenomena.
Work connects the force applied on an object and the displacement it undergoes. It quantifies how much energy is transferred to or from the object by that force.
Energy is the capacity of a body to do work. It exists in various forms, such as kinetic energy (energy of motion) and potential energy (stored energy due to position or configuration).
Understanding the relationship between work and energy allows us to solve many physics problems efficiently, especially in competitive exams where time and clarity matter. This section will build these concepts step-by-step, starting from the definition of work, moving to energy forms, and culminating in the powerful work-energy theorem.
When a force acts on an object causing it to move, the force may transfer energy to the object. This transfer is called work. But how do we measure work?
Work is defined as the dot product (or scalar product) of the force vector \(\vec{F}\) and the displacement vector \(\vec{d}\). This means:
The mathematical expression for work done by a constant force is:
Here, \(\theta\) is the angle between the force and displacement directions.
Interpretation of work sign:
Work Done by Variable Force: When the force varies with position, work is calculated by integrating the force along the path:
\[W = \int_{\text{initial}}^{\text{final}} \vec{F} \cdot d\vec{r}\]This integral sums the infinitesimal work done over small displacements where force can be considered constant.
Objects in motion possess energy due to their movement, called kinetic energy. It depends on the mass of the object and the speed at which it moves.
To derive the formula for kinetic energy, consider a body of mass \(m\) initially moving at velocity \(v_i\). A net force \(\vec{F}\) acts on it over a displacement \(d\), increasing its velocity to \(v_f\).
From Newton's second law, \(\vec{F} = m \vec{a}\), and using the work done by the force:
\[W = F d = m a d\]Using the kinematic relation \(v_f^2 = v_i^2 + 2 a d\), we solve for \(a d\):
\[a d = \frac{v_f^2 - v_i^2}{2}\]Substituting back into work expression:
\[W = m \times \frac{v_f^2 - v_i^2}{2} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2\]This shows that the work done by the net force equals the change in kinetic energy:
While kinetic energy is energy of motion, potential energy is energy stored due to an object's position or configuration. It represents the capacity to do work because of the object's position in a force field or due to deformation.
Near Earth's surface, an object of mass \(m\) raised to a height \(h\) above a reference level (usually the ground) has gravitational potential energy given by:
The zero of potential energy is arbitrary; it depends on the chosen reference level. Usually, the ground or floor is taken as zero potential energy.
A stretched or compressed spring stores elastic potential energy. For a spring with spring constant \(k\), stretched or compressed by displacement \(x\) from its equilibrium length, the potential energy stored is:
The work-energy theorem is a powerful principle that connects the net work done on a body to its change in kinetic energy. It states:
Mathematically,
\[W_{net} = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2\]This means that if you know the total work done by all forces acting on an object, you can find how its speed changes without directly calculating forces or acceleration.
graph TD F[Net Force] --> W[Work Done] W --> K[Change in Kinetic Energy]
Power measures how fast work is done or energy is transferred. It is the rate of doing work:
\[P = \frac{W}{t}\]where \(W\) is work done in time \(t\).
Power can also be expressed in terms of force and velocity vectors as:
\[P = \vec{F} \cdot \vec{v}\]This formula shows that power depends on the component of force in the direction of velocity.
Average power is the total work done divided by total time, while instantaneous power is the power at a specific instant.
The SI unit of power is the watt (W), where 1 W = 1 joule/second.
Step 1: Identify given data:
Step 2: Use the formula for work done by a constant force:
\[ W = F d \cos \theta \]Step 3: Calculate \(\cos 60^\circ = 0.5\).
Step 4: Substitute values:
\[ W = 10 \times 5 \times 0.5 = 25\, \text{J} \]Answer: The work done by the force is 25 joules.
Step 1: Given:
Step 2: Calculate work done by the force:
\[ W = F \times d = 20 \times 3 = 60\, \text{J} \]Step 3: Apply work-energy theorem:
\[ W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 = \frac{1}{2} m v_f^2 \]Step 4: Solve for \(v_f\):
\[ 60 = \frac{1}{2} \times 2 \times v_f^2 \implies 60 = v_f^2 \implies v_f = \sqrt{60} \approx 7.75\, \text{m/s} \]Answer: The final velocity of the block is approximately 7.75 m/s.
Step 1: Given:
Step 2: Use conservation of mechanical energy:
\[ \text{Potential energy at top} = \text{Kinetic energy at bottom} \] \[ m g h = \frac{1}{2} m v_f^2 \]Step 3: Cancel \(m\) and solve for \(v_f\):
\[ v_f = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 20} = \sqrt{392} \approx 19.8\, \text{m/s} \]Answer: The velocity just before impact is approximately 19.8 m/s.
Step 1: Given:
Step 2: The force exerted by the spring is variable and given by Hooke's law:
\[ F = -k x \]Step 3: Work done to stretch the spring from 0 to \(x\) is:
\[ W = \int_0^x F \, dx = \int_0^x k x \, dx = \frac{1}{2} k x^2 \]Step 4: Substitute values:
\[ W = \frac{1}{2} \times 200 \times (0.1)^2 = 0.5 \times 200 \times 0.01 = 1\, \text{J} \]Answer: The work done in stretching the spring is 1 joule.
Step 1: Given:
Step 2: Force required to lift the load at constant speed (equal to weight):
\[ F = m g = 50 \times 9.8 = 490\, \text{N} \]Step 3: Power is:
\[ P = F \times v = 490 \times 2 = 980\, \text{W} \]Answer: The motor develops 980 watts (or 0.98 kW) of power.
When to use: When the force is applied at an angle to the displacement.
When to use: When net work and velocity changes are involved, especially on frictionless surfaces.
When to use: When force direction is perpendicular to motion, such as centripetal force in circular motion.
When to use: When dealing with variable spring forces or calculating work done in stretching/compressing springs.
When to use: Throughout problem solving to ensure correct numerical answers.
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