NEET UG - Work energy theorem, power, potential energy of spring E-Book Study Material

Work, Energy, and Power

Work Energy Theorem

The Work-Energy Theorem is a fundamental principle in physics that relates the work done on an object to the change in its kinetic energy. It states that the net work done on a particle by all forces acting on it is equal to the change in the kinetic energy of that particle. This theorem provides a powerful link between the concepts of work and energy.

Mathematically, the Work-Energy Theorem is expressed as:

$W_{net} = \Delta KE$

Where:

$W_{net}$ is the net work done on the object.
$\Delta KE$ is the change in the kinetic energy of the object.

The change in kinetic energy is calculated as the final kinetic energy minus the initial kinetic energy:

$\Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Where:

$m$ is the mass of the object.
$v_f$ is the final velocity of the object.
$v_i$ is the initial velocity of the object.

Let's understand how this theorem is derived. Consider a particle of mass $m$ moving with an initial velocity $v_i$. If a constant net force $F_{net}$ acts on it over a displacement $d$, the particle's velocity changes to $v_f$. According to Newton's second law, $F_{net} = ma$, where $a$ is the acceleration.

We can use the kinematic equation: $v_f^2 = v_i^2 + 2ad$. Rearranging this equation to solve for $ad$:

$ad = \frac{v_f^2 - v_i^2}{2}$

Now, let's consider the work done by the net force. Work done by a constant force is given by $W = F \cdot d$. For motion in one dimension, $W_{net} = F_{net} \cdot d$.

Substitute $F_{net} = ma$:

$W_{net} = (ma)d = m(ad)$

Now substitute the expression for $ad$:

$W_{net} = m \left( \frac{v_f^2 - v_i^2}{2} \right)$

$W_{net} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

This is exactly the change in kinetic energy, $\Delta KE$. Thus, $W_{net} = \Delta KE$.

Key Points about Work-Energy Theorem:

It applies to the *net* work done by *all* forces.
Work done by individual forces contributes to the total change in kinetic energy.
If the net work done is positive ($W_{net} > 0$), the kinetic energy increases, and the object speeds up.
If the net work done is negative ($W_{net} < 0$), the kinetic energy decreases, and the object slows down.
If the net work done is zero ($W_{net} = 0$), the kinetic energy remains constant, meaning the velocity does not change.

Example: A block of mass 2 kg is initially at rest. A net force of 10 N acts on it for 5 meters. What is its final kinetic energy and final velocity?

Given: $m = 2$ kg, $v_i = 0$ m/s (at rest), $F_{net} = 10$ N, $d = 5$ m.

Work done by the net force: $W_{net} = F_{net} \times d = 10 \text{ N} \times 5 \text{ m} = 50 \text{ J}$.

According to the Work-Energy Theorem, $W_{net} = \Delta KE$.

So, $\Delta KE = 50$ J.

Since the initial kinetic energy $KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(2 \text{ kg})(0 \text{ m/s})^2 = 0$ J.

The final kinetic energy is $KE_f = KE_i + W_{net} = 0 \text{ J} + 50 \text{ J} = 50 \text{ J}$.

Now, to find the final velocity: $KE_f = \frac{1}{2}mv_f^2$.

$50 \text{ J} = \frac{1}{2}(2 \text{ kg})v_f^2$

$50 = v_f^2$

$v_f = \sqrt{50} = 5\sqrt{2}$ m/s.

Power

Power is the rate at which work is done or energy is transferred. It measures how quickly work is performed. A machine or a person that can do a large amount of work in a short time is said to have high power.

The average power ($P_{avg}$) is defined as the total work done ($W$) divided by the time interval ($\Delta t$) over which the work is done:

$P_{avg} = \frac{W}{\Delta t}$

The SI unit of power is the Watt (W), named after James Watt. One Watt is equal to one Joule per second (1 W = 1 J/s).

Instantaneous power ($P$) is the power at a specific moment in time. It is the time derivative of work:

$P = \frac{dW}{dt}$

Since work done by a force $F$ over an infinitesimal displacement $ds$ is $dW = \vec{F} \cdot d\vec{s}$, we can write instantaneous power as:

$P = \frac{\vec{F} \cdot d\vec{s}}{dt} = \vec{F} \cdot \frac{d\vec{s}}{dt}$

Recognizing that $\frac{d\vec{s}}{dt}$ is the instantaneous velocity $\vec{v}$, we get another important formula for instantaneous power:

$P = \vec{F} \cdot \vec{v}$

This formula is particularly useful when dealing with forces and velocities. It shows that power is the dot product of the force vector and the velocity vector. If the force and velocity are in the same direction, $P = Fv$. If they are in opposite directions, $P = -Fv$. If they are perpendicular, the power is zero ($P=0$), meaning the force is doing no work at that instant.

Units of Power:

SI Unit: Watt (W)
Other common units:

Kilowatt (kW): $1 \text{ kW} = 1000 \text{ W}$
Megawatt (MW): $1 \text{ MW} = 10^6 \text{ W}$
Horsepower (hp): $1 \text{ hp} \approx 746 \text{ W}$ (This is an imperial unit, often used for engines)

Example: A motor lifts a 10 kg load vertically by 5 meters in 2 seconds. Calculate the average power of the motor. (Assume $g = 10 \text{ m/s}^2$)

First, calculate the work done by the motor. The force required to lift the load against gravity is equal to its weight, $F = mg$.

$F = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100 \text{ N}$.

The work done is $W = F \times d = 100 \text{ N} \times 5 \text{ m} = 500 \text{ J}$.

The time taken is $\Delta t = 2$ s.

The average power is $P_{avg} = \frac{W}{\Delta t} = \frac{500 \text{ J}}{2 \text{ s}} = 250 \text{ W}$.

Example 2: A car engine delivers a force of 3000 N to move the car forward at a speed of 20 m/s. What is the power output of the engine at this instant?

Given: $F = 3000$ N, $v = 20$ m/s. The force and velocity are in the same direction.

Using the formula $P = Fv$:

$P = 3000 \text{ N} \times 20 \text{ m/s} = 60000 \text{ W}$.

This can also be expressed as 60 kW.

Potential Energy of a Spring

A spring is an example of a system that stores potential energy due to its deformation (stretching or compression). The force exerted by an ideal spring is called a restoring force because it always acts in a direction opposite to the displacement from its equilibrium position. This force is described by Hooke's Law.

Hooke's Law: Hooke's Law states that the force ($F_s$) exerted by a spring is directly proportional to the displacement ($x$) from its equilibrium position and acts in the opposite direction.

$F_s = -kx$

Where:

$F_s$ is the restoring force exerted by the spring.
$k$ is the spring constant, a measure of the stiffness of the spring. A higher $k$ means a stiffer spring. Its SI unit is N/m.
$x$ is the displacement of the spring from its equilibrium position.
The negative sign indicates that the restoring force is always opposite to the displacement.

To stretch or compress a spring, an external force ($F_{ext}$) must be applied, which is equal in magnitude and opposite in direction to the restoring force of the spring:

$F_{ext} = -F_s = -(-kx) = kx$

This external force does work on the spring, and this work is stored as potential energy in the spring.

Calculating the Potential Energy of a Spring: The potential energy stored in a spring ($U_s$) is the work done by the external force to displace the spring from its equilibrium position ($x=0$) to a displacement $x$. We can calculate this work by integrating the external force over the displacement.

$W_{ext} = \int_{0}^{x} F_{ext} dx = \int_{0}^{x} (kx) dx$

$W_{ext} = k \int_{0}^{x} x dx = k \left[ \frac{x^2}{2} \right]_{0}^{x} = k \left( \frac{x^2}{2} - \frac{0^2}{2} \right)$

$W_{ext} = \frac{1}{2}kx^2$

This work done is stored as potential energy in the spring. Therefore, the potential energy of a spring is given by:

$U_s = \frac{1}{2}kx^2$

Where:

$U_s$ is the elastic potential energy stored in the spring.
$k$ is the spring constant.
$x$ is the displacement from the equilibrium position.

Important Considerations:

The potential energy is zero when the spring is at its equilibrium position ($x=0$).
The potential energy is always positive, regardless of whether the spring is stretched ($x>0$) or compressed ($x<0$), because $x^2$ is always non-negative.
The potential energy is proportional to the square of the displacement. Doubling the displacement quadruples the potential energy.

Example: A spring has a spring constant $k = 200$ N/m. (a) How much work must be done to stretch the spring 0.1 meters from its equilibrium position? (b) How much work must be done to compress the spring 0.1 meters from its equilibrium position? (c) What is the potential energy stored in the spring when it is stretched by 0.1 meters?

Given: $k = 200$ N/m.

(a) Work to stretch by $x = 0.1$ m:

$W_{stretch} = \frac{1}{2}kx^2 = \frac{1}{2}(200 \text{ N/m})(0.1 \text{ m})^2$

$W_{stretch} = \frac{1}{2}(200)(0.01) \text{ J} = 100 \times 0.01 \text{ J} = 1 \text{ J}$.

(b) Work to compress by $x = -0.1$ m (displacement is -0.1 m from equilibrium):

$W_{compress} = \frac{1}{2}kx^2 = \frac{1}{2}(200 \text{ N/m})(-0.1 \text{ m})^2$

$W_{compress} = \frac{1}{2}(200)(0.01) \text{ J} = 1 \text{ J}$.

Note that the work done is the same for stretching and compressing by the same distance.

(c) Potential energy stored when stretched by $x = 0.1$ m:

$U_s = \frac{1}{2}kx^2 = \frac{1}{2}(200 \text{ N/m})(0.1 \text{ m})^2 = 1 \text{ J}$.

Conservation of Mechanical Energy with Springs: In a system involving a spring and other conservative forces (like gravity), the total mechanical energy (sum of kinetic energy and potential energy) is conserved if only conservative forces do work.

$E_{total} = KE + PE_{gravity} + PE_{spring} = \text{constant}$

This means that energy can transform between kinetic energy, gravitational potential energy, and spring potential energy, but their sum remains constant.

Example: Spring-Mass System Oscillations Consider a mass $m$ attached to a spring with spring constant $k$. If the mass is displaced from equilibrium and released, it will oscillate. At the maximum displacement ($x = \pm A$, where $A$ is the amplitude), the velocity is zero ($KE=0$), and all the energy is stored as spring potential energy: $E_{total} = \frac{1}{2}kA^2$. At the equilibrium position ($x=0$), the potential energy is zero ($U_s=0$), and all the energy is kinetic energy: $E_{total} = \frac{1}{2}mv_{max}^2$. By conservation of energy:

$\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$

This relationship is crucial for understanding simple harmonic motion.