JEE Physics Work Energy and Power Notes Set 01

Work Done by a Force

When a force moves its point of application, the force is said to do work. For a constant force the work is the scalar (dot) product of force and displacement: \( W = \vec F\cdot\vec d = Fd\cos\theta \), where \( \theta \) is the angle between them. Work is a scalar, measured in joule (1 J = 1 N m = 1 kg m²/s²). Its dimensional formula is \( [ML^2T^{-2}] \).
Other units: 1 erg = \( 10^{-7} \) J, 1 eV = \( 1.6\times 10^{-19} \) J, 1 kWh = \( 3.6\times 10^{6} \) J.

Sign of Work

Positive work: \( 0^\circ \le \theta < 90^\circ \), the force helps the motion (gravity on a falling body).
Zero work: the force is zero, or the displacement is zero, or \( \theta = 90^\circ \). Examples are the tension on a pendulum bob, the normal force on a sliding block, and the centripetal force on an orbiting satellite.
Negative work: \( 90^\circ < \theta \le 180^\circ \), the force opposes the motion (friction, braking force, gravity on a rising body).

Work by a Variable Force

If the force changes, divide the path into tiny elements \( d\vec r \) over which the force is nearly constant, and add up: \( W = \int \vec F\cdot d\vec r \). In components, \( W = \int (F_x\,dx + F_y\,dy + F_z\,dz) \). For a force along a straight line, the work equals the area under the force-displacement graph; areas below the axis count as negative.
Work depends on the frame of reference because displacement depends on the frame, while the force does not.

Kinetic Energy and the Work-Energy Theorem

The energy a body possesses because of its motion is \( K = \tfrac{1}{2}mv^2 \). It is never negative and depends on the frame of reference. In terms of momentum, \( K = \dfrac{p^2}{2m} \), which gives the useful results: for equal momenta, K is inversely proportional to m; for equal energies, p is proportional to \( \sqrt{m} \).
Work-energy theorem: the net work done by all forces (conservative or not) on a particle equals its change in kinetic energy, \( W_{net} = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2 \). For a variable force it follows from \( W = \int m\,\dfrac{dv}{dt}\,dx = \int mv\,dv \).

How to Apply the Theorem

1. Mark the initial and final positions and write the kinetic energy at each.
2. Draw the free-body diagram and find the work done by each force (with sign).
3. Add the works and equate the sum to the change in kinetic energy.
Use the theorem whenever the question gives speeds and distances but not time.

Worked Examples

Worked Example 1. A force \( F = (4 + 3x) \) N acts on a body along the x-axis. Find the work done in moving it from x = 1 m to x = 3 m.
Answer:
\( W = \int_1^3 (4 + 3x)\,dx = \left[4x + \tfrac{3}{2}x^2\right]_1^3 \).
\( = (12 + 13.5) - (4 + 1.5) = 20 \) J.

Worked Example 2. A 2 kg body moving at 3 m/s is acted on by a net force that raises its speed to 7 m/s. Find the work done.
Answer:
\( W = \tfrac{1}{2}(2)(7^2 - 3^2) = 40 \) J.

Worked Example 3. A bullet of mass 20 g moving at 300 m/s penetrates a wooden block to a depth of 15 cm. Find the average resisting force.
Answer:
Loss of KE \( = \tfrac{1}{2}(0.02)(300)^2 = 900 \) J.
\( F \times 0.15 = 900 \Rightarrow F = 6000 \) N.