Work, Power and Energy for Engineering Physics

Work:
Work is said to be done when a force acts on a body and the body is displaced in the direction of the force.
SI Unit: Joule (J)

Power:
Power is the rate at which work is done or energy is transferred.
SI Unit: Watt (W)

Energy:
Energy is the capacity of a body to do work.
SI Unit: Joule (J)

Kinetic Energy:
Kinetic energy is the energy possessed by a body due to its motion. Any object that is moving has kinetic energy.

Potential Energy:
Potential energy is the energy stored in a body due to its position or configuration. For example, an object raised to a height has gravitational potential energy.

Consider a body of mass m initially at rest. A force is applied on it, causing it to gain a velocity v.

Let the body move a distance s under the action of the constant force.

Work done on the body:

W = F × s

From Newton’s second law:

F = m a

Using the equation of motion:

v² = u² + 2as

Since the body starts from rest, u = 0:

v² = 2as ⇒ s = v² / (2a)

Now substitute F = ma and s = v² / (2a):

W = ma × (v² / 2a)

After cancelling a:

W = (1/2) m v²

Thus, the kinetic energy of a body is:

KE = (1/2) m v²

The momentum of a body is given by:

p = mv

The kinetic energy of a body is:

KE = (1/2) m v²

Substitute v = p/m in the kinetic energy formula:

KE = (1/2) m (p/m)²

Simplifying:

KE = (1/2) m × (p² / m²)

KE = p² / (2m)

Thus, the relation between kinetic energy and momentum is:

KE = p² / (2m)

Principle of Conservation of Mechanical Energy:
The total mechanical energy of an isolated system remains constant if only conservative forces (such as gravitational or elastic forces) act on the system.

This means:

Total Mechanical Energy = Kinetic Energy + Potential Energy = Constant

Mechanical energy may change from kinetic to potential or vice versa, but the sum remains unchanged.

Consider a body of mass m falling freely from a height h above the ground.

1. At the top (initial position)

Velocity v = 0

Kinetic Energy: KE = 0

Potential Energy: PE = mgh

Total Energy: E = mgh

2. At some point during the fall

Let the body fall through a distance x.

Height above ground = (h − x)

Potential Energy:

PE = mg(h − x)

Using equation of motion:

v² = 2gx

Kinetic Energy:

KE = (1/2)mv² = (1/2)m(2gx) = mgx

Total Energy:

E = KE + PE = mgx + mg(h − x) = mgh

3. Just before touching the ground

Height = 0

Potential Energy: PE = 0

Velocity: v² = 2gh

Kinetic Energy: KE = (1/2)m(2gh) = mgh

Total Energy: E = mgh

Conclusion:

At every position during the fall, the total mechanical energy is:

E = mgh = constant

Hence, the total energy of a freely falling body remains constant.

Work-Energy Principle:
The work done by the net force acting on a body is equal to the change in its kinetic energy.

Mathematically,

Work done (W) = Final Kinetic Energy − Initial Kinetic Energy

W = (1/2)mv_f^2 − (1/2)mv_i^2

This principle connects force, displacement, and the resulting change in motion of a body.

Given:

Force, F = 30 N
Displacement, d = 5 m

Work done is given by:

W = F × d

Substituting values:

W = 30 × 5 = 150 J

Work done = 150 Joules

Given:

Force, F = 5000 N
Distance, d = 200 m

Work done is:

W = F × d

Substituting values:

W = 5000 × 200 = 1,000,000 J

Work done = 1,000,000 Joules (or 1 MJ)

Given:

Force, F = 100 N
Distance moved, d = 8 m
Angle between force and displacement, θ = 60°

Work done is given by:

W = F × d × cosθ

Substituting values:

W = 100 × 8 × cos60°

cos60° = 0.5

W = 800 × 0.5 = 400 J

Work done = 400 Joules

Given:

Weight (force), F = 80 N
Distance lifted, d = 5 m
Time taken, t = 4 s

First calculate the work done:

W = F × d = 80 × 5 = 400 J

Power is given by:

P = W / t

Substituting:

P = 400 / 4 = 100 W

Power = 100 Watts

Given:

Mass, m = 500 kg
Velocity, v = 25 m/s

Kinetic energy is given by:

KE = (1/2) m v²

Substituting values:

KE = (1/2) × 500 × (25)²

KE = 250 × 625 = 156250 J

Kinetic Energy = 156,250 Joules

Given:

Mass of block, m = 2 kg
Incline angle, θ = 30°

The block starts from rest, so its initial kinetic energy is zero.

On a frictionless incline, loss of potential energy = gain in kinetic energy.

If the block moves down a vertical height h:

mgh = (1/2)mv²

Mass cancels out:

gh = (1/2)v²

v = √(2gh)

Since the incline is 30°, the vertical height is related to the length of the incline L by:

h = L sin30° = L × 1/2

Substitute h = L/2 into the energy equation:

v = √(2g × L/2)

v = √(gL)

Taking g = 9.8 m/s² and assuming the incline has length L (cancels out in the ratio), the final expression becomes:

v = √(gL)

But for a 30° frictionless incline, the standard result is:

v = √(gh) where h is vertical drop

Thus for any vertical height h:

v = √(2gh)

Without a specific height or length given, the final answer is expressed in terms of height:

Final Speed: v = √(2gh)

If you give the height or length of the incline, I can calculate the exact numerical value.

Given:

Force, F = 50 N
Distance, d = 10 m

1. Work Done

Since the force is applied in the direction of motion:

W = F × d

Substituting values:

W = 50 × 10 = 500 J

Work Done = 500 Joules

2. Final Kinetic Energy

If the crate starts from rest, the work done on it becomes its kinetic energy (Work–Energy Theorem):

KE = W = 500 J

Final Kinetic Energy = 500 Joules