Work–Energy Theorem

The Work-Energy Theorem is an important concept of physics that helps in understanding how forces affect motion by using the relation of work done by forces to changes in an object's kinetic energy.

1.0Insight Into Work-Energy Theorem

According to the Work-Energy Theorem, the net force that acts on an object does work on the object equal to the change in its kinetic energy. Mathematically, it can be stated as:

$W = Δ K = K_{f} - K_{i}$

Here:

W is the work done on the object,
K_f is the final kinetic energy of the object,
K_i is the initial kinetic energy of the object.

2.0Breaking the Components of Work-Energy Theorem

Work Done

Work (W) is defined as the dot product of force (F) and displacement(d). Where force is the net force applied to an object and d is the net distance covered by the object due to applied force. Mathematically, it can be expressed as:

$W = F . d = F d cos θ$

Here $θ$ is the angle between the force and the displacement vector. This formula is used for the constant force.

For Variable force, work done can be calculated as:

$W =^{^{r f}} \int_{r i} F . d r$

Here, r_i and r_f are the initial and final positions of the object, respectively.

Kinetic Energy

The Kinetic Energy (K) of an object of mass “m” is the energy possessed by any object due to its motion and is moving with a velocity “v”. It can be calculated by the formula:

$K = \frac{1}{2} m v^{2}$

The change in kinetic energy is:

$Δ K = K_{f} - K_{i} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$

Where v_f = final velocity and v_i = initial velocity of the object.

3.0Derivation of Work-Energy Theorem

Imagine an object in rectilinear motion under the influence of constant acceleration a, with initial velocity being v_i and final velocity v_f and it travelled a distance “d”.

To Prove: $W = K_{f} - K_{i}$

Derivation:

Using the third equation of motion,

$v^{2} - u^{2} = 2 a s$

Here,

u = initial speed of an object

v = final speed of an object

s = distance traversed by that object

Equating the values in the above equation,

$v_{f}^{2} - u_{i}^{2} = 2 a s$ …….(1)

Multiply equation 1 by m/2 on both sides,

$\frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = 2 \times \frac{m}{2} \times a s$

$\frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = ma s$ ……..(2)

From Newton’s Second law of motion, we know that force F = ma

Hence, equation 2 will become,

$\frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = F s$

We know that, $K = \frac{1}{2} m v^{2}$ and Work done W = Fs

Hence,

$K_{f} - K_{i} = W$

4.0Application of Work-Energy Theorem

The Work-Energy theorem can be used to solve many problems related to different concepts of physics as well as real-life problems.

Kinetic Energy Calculation: It will determine the change in kinetic energy of an object due to work done by forces, which further helps in speed and motion analysis.
Motion Analysis: The theorem is important to understand how forces affect the velocity and acceleration of objects, either accelerating or decelerating.
Energy Conservation: It tracks the conversion between kinetic and potential energy, especially in frictionless systems such as free-falling objects or projectiles.
Collision Problems: It is involved in the understanding of energy transfer due to collisions as it helps solve the difference between elastic and inelastic collisions by its relation to a change in kinetic energy.
Plane Inclines: It aids in solving questions that involve motions on inclined planes where gravity, a normal force, and friction are of concern.
Varying Forces: The theorem is applicable to variable forces since integration can easily calculate work.
Power: it calculates the rate of energy transfer or work done over time, in other words, power in mechanical systems.

5.0Solved Examples

Problem 1: A particle of mass m = 2 kg is moving along the xxx-axis under the influence of a force F(x) = 4x, where x is the position of the particle in meters. The particle starts at rest at x=0 and moves to x = 5 m. Using the Work-Energy Theorem, calculate the final velocity of the particle at x = 5 m.

Solution: Using the formula,

$W =^{^{x f}} \int_{x i} F (x) . d x =^{^{5}} \int_{0} 4 x d x$

$W = [\frac{x ^{2}}{2}]_{0}^{5} = 4 (\frac{25}{2} - 0) = 50 J$

Using the Work-Energy Theorem:

$W = K_{f} - K_{i}$

$50 = K_{f} - 0$

Here, K_i = 0 as the initial velocity is 0.

The kinetic energy of the particle:

$K = \frac{1}{2} m v^{2}$

$50 = \frac{1}{2} \times 2 \times v^{2}$

$v^{2} = 50$

$v = 50 = 7.07 m / s$

Problem 2: A 5 kg block is pushed along a horizontal surface with a constant force of 30 N. The coefficient of kinetic friction between the block and the surface is μ_k=0.2, and the block moves a distance of 10 meters. Find the final velocity of the block if it starts from rest.

Solution: the frictional force is given by:

$F_{f r i c t i o n} = μ_{k} N$

Here, N = mg is the normal force. So, the frictional force:

$F_{f r i c t i o n} = 0.2 \times 5 \times 9.8 = 9.8 N$

The net force will be

$F_{n e t} = F_{a ppl i e d} - F_{f r i c t i o n} = 30-9.8 = 20.2 N$

Work done by the net force will be

$W = F_{n e t} . d = 20.2 \times 10 = 202 J$

Using the work-energy theorem:

$W = K_{f} - K_{i}$

$202 = \frac{1}{2} m v^{2} - 0$

$202 = \frac{1}{2} \times 5 v^{2}$

$v^{2} = \frac{202}{2.5} = 80.8$

$v = 80.8 = 8.99 m / s$

Problem 3: A 3 kg object is lifted vertically from the ground to a height of 5 meters. Determine the work done on the object and its velocity if it is released from rest and falls under gravity.

Solution: Work done by gravity

$W = F_{g r a v i t y} h$

$W = 29.4 \times 5 = 147 J$

Using the Work-Energy theorem:

$W = K_{f} - K_{i}$

$147 = \frac{1}{2} m v^{2} - 0$

$147 = \frac{1}{2} \times 3 v^{2}$

$v^{2} = 49 \times 2 = 98$

$v = 98 = 9.9 m / s$