Work-Energy Theorem by Constant Force: Comprehensive NEET Physics Notes

1. Work-Energy Theorem

The work-energy theorem connects the work done by forces acting on a particle to the change in its kinetic energy. It plays a crucial role in analyzing motion when constant forces are involved. The theorem states that the net work done by forces acting on an object is equal to the change in its kinetic energy.

1.1 Statement of Work-Energy Theorem

The work-energy theorem can be summarized mathematically as:

$W_{net} = Δ K = K_{f} - K_{i}$

Where:

$W_{net}$ is the net work done,
$K_{i}$ is the initial kinetic energy,
$K_{f}$ is the final kinetic energy.

This relation highlights how the total work done on an object results in either an increase or decrease in its kinetic energy, depending on the direction and magnitude of the forces involved.

1.2 Derivation of the Work-Energy Theorem for Constant Force

Consider a particle of mass $m$ subjected to a constant force $f$ . By Newton’s second law:

$F = ma$

Where:

$a$ is the constant acceleration of the object,
$F$ is the constant applied force.

The work done, $W$ , by the force during a displacement from an initial position $x_{i}$ to a final position $x_{f}$ is given by:

$W = F \cdot d = F (x_{f} - x_{i})$

Using the third equation of motion:

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

Where:

$v$ is the final velocity,
$u$ is the initial velocity,
$s$ is the displacement.

Multiplying by $\frac{m}{2}$ :

$\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = ma s$

Since $F = ma$ , we substitute:

$\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = F \cdot s$

Thus, the work done by the force equals the change in kinetic energy:

$W = Δ K = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

This is the work-energy theorem for constant forces, showing that the work done by a force results in a change in the kinetic energy of the object.

2. Applications and Special Cases

2.1 Work Done by Gravity

When an object moves under the influence of gravitational force, the work-energy theorem simplifies to:

$W = m g h$

Where:

$m$ is the mass of the object,
$g$ is the gravitational acceleration,
$h$ is the vertical displacement.

Real-Life Application:

This application is seen in activities like lifting objects. For instance, when you lift a box, the work done against gravity can be calculated using this formula.

2.2 Work Done by Friction

In cases where friction is involved, the work-energy theorem accounts for the negative work done by the frictional force. If $f$ is the frictional force and $s$ is the displacement:

$W_{f} = - f \cdot s$

Since friction opposes motion, it results in a reduction of the object’s kinetic energy.

NEET Tip:

In NEET problems involving friction, remember that the work done by friction is always negative and reduces the object’s total energy.

3. Visual Aids

To enhance understanding, diagrams can illustrate how forces and displacements interact in real scenarios. A diagram showing the force applied at an angle with displacement would visually demonstrate the concept of work. Diagrams representing the difference in kinetic energies at various points during motion will also clarify the principle.

Quick Recap

The work-energy theorem relates net work to changes in kinetic energy.
For a constant force, work is the product of force and displacement.
In real-life applications, the theorem simplifies calculations, like lifting objects or calculating the effect of friction.
Work done by non-conservative forces like friction is negative and reduces total energy.

NEET Problem-Solving Strategy

When approaching NEET problems using the work-energy theorem:
Identify the forces acting on the object and determine if they are conservative or non-conservative.
Calculate the work done using $W = F \cdot s$ , accounting for the direction of forces.
Use the work-energy theorem to relate the work done to the change in kinetic energy.

Practice Questions

A block of mass 3 kg is pushed with a force of 12 N across a smooth surface for 4 m. Calculate the work done and the final velocity if the initial velocity was 0 m/s.
A 10 kg object is lifted to a height of 15 m. Find the work done against gravity and the final kinetic energy if the object started at rest.
A car of mass 1000 kg comes to a stop from 30 m/s due to a braking force of 4000 N. Using the work-energy theorem, find the stopping distance.
A box is dragged across a surface with a constant frictional force of 60 N for 10 m. Calculate the work done by friction.
A particle initially moving at 8 m/s is acted upon by a force that increases its velocity to 16 m/s. If the particle has a mass of 2 kg, calculate the work done.

Solutions to Practice Questions

Work done: $W = F \cdot s = 12 \times 4 = 48, J$ .
Final velocity: $v = \frac{2 W}{m} = \frac{2 \times 48}{3} = 8, m/s$ .
Work done by gravity: $W = m g h = 10 \times 9.8 \times 15 = 1470, J$ .
Final kinetic energy: $K = W = 1470, J$ .
Work-energy theorem: $W = \frac{1}{2} m v^{2} = - 4000 \times d$ . Solving for $d$ gives the stopping distance.
Work done by friction: $W_{f} = - 60 \times 10 = - 600 J$ .
Work done: $Δ K = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = \frac{1}{2} \times 2 \times$