In Physics, the term 'work' is used only when a force produces displacement in a body.
Definition: Work is said to be done only when the force applied on a body makes the body move.
Fig 2.1 A man pushing a car
If there is no displacement (e.g., pushing a rigid wall or holding a heavy load while standing still), the work done is zero.
Where $F$ is force, $S$ is displacement, and $\theta$ is the angle between them.
(i) $\theta = 0^\circ$ (Positive Work): Force & displacement in same direction. $W = F \times S$.
(ii) $\theta = 90^\circ$ (Zero Work): Force & displacement are perpendicular. $W = 0$ (e.g., Circular orbit).
(iii) $\theta = 180^\circ$ (Negative Work): Displacement opposite to force. $W = -F \times S$ (e.g., Friction).
The area under a Force-Displacement graph gives the work done.
where $h$ is the vertical height climbed or descended.
Power is the rate of doing work.
| Work | Power |
|---|---|
| Force $\times$ Displacement. | Rate of doing work. |
| Independent of time. | Depends on time. |
A crane pulls up a car of mass $500 \text{ kg}$ to a vertical height of $4 \text{ m}$ in $20 \text{ s}$. Calculate power. ($g = 9.8$)
Solution:
$W = F \times h = (500 \times 9.8) \times 4 = 19600 \text{ J}$.
$P = W/t = 19600 / 20 = 980 \text{ W}$.
Definition: Energy is the capacity to do work. It is a scalar quantity with the same units as work (Joules/Ergs).
In various devices, energy is converted from one form to another. Examples:
The energy possessed by a body due to its changed position or configuration.
The energy possessed by a body due to its state of motion.
Work done by a force on a moving body equals the increase in its kinetic energy: $W = K_f - K_i = \Delta K$.
Definition: Energy can neither be created nor destroyed; it only changes form. Total mechanical energy ($K + U$) remains constant in absence of friction.
For a freely falling body, total energy $E = mgh$ at all points.
| Position | Kinetic (K) | Potential (U) | Total (E) |
|---|---|---|---|
| A (Highest) | 0 | $mgh$ | $mgh$ |
| B (Middle) | $\frac{1}{2}mgh$ | $\frac{1}{2}mgh$ | $mgh$ |
| C (Ground) | $mgh$ | 0 | $mgh$ |
A boy of mass $40 \text{ kg}$ climbs $8 \text{ m}$ in $5 \text{ s}$. Power? ($g=10$)
Ans: $P = \frac{mgh}{t} = \frac{40 \times 10 \times 8}{5} = 640 \text{ W}$.
In a dam, water falls at $1000 \text{ kg/s}$ from $100 \text{ m}$. Calculate power if efficiency is $60\%$.
Ans: 1s Potential Energy = $mgh = 1000 \times 10 \times 100 = 10^6 \text{ J}$.
Power Generated = $60\% \text{ of } 10^6 = 600,000 \text{ W} = 600 \text{ kW}$.
A block of mass $30 \text{ kg}$ is pulled up a slope of length $3 \text{ m}$ and height $1.5 \text{ m}$ with force $200 \text{ N}$. Calculate: (i) Work done, (ii) Potential energy gain.
Ans: (i) $W = F \times \text{length} = 200 \times 3 = 600 \text{ J}$.
(ii) $U = mgh = 30 \times 10 \times 1.5 = 450 \text{ J}$. (The $150 \text{ J}$ difference is lost to friction).
A $10\text{ g}$ ball falls from $5\text{ m}$ and rebounds to $4\text{ m}$. Find loss in energy. ($g=9.8$)
Ans: Initial PE = $0.01 \times 9.8 \times 5 = 0.49\text{ J}$.
Final PE = $0.01 \times 9.8 \times 4 = 0.392\text{ J}$.
$\text{Loss} = 0.49 - 0.392 = 0.098\text{ J}$ (Lost as heat/sound).
A skier ($60 \text{ kg}$) takes off from A ($75 \text{ m}$) to B ($15 \text{ m}$). Calculate speed at B if $75\%$ energy becomes KE.
Ans: $\Delta U = mg(h_1 - h_2) = 60 \times 10 \times (60) = 36000 \text{ J}$.
$K = 0.75 \times 36000 = 27000 \text{ J}$.
$\frac{1}{2}mv^2 = 27000 \implies \frac{1}{2}(60)v^2 = 27000 \implies v^2 = 900 \implies v = 30 \text{ m/s}$.
Practice makes perfect. Review the numericals thrice.
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