Work and Power

⚛️ Work as an Integral

Part 1 of 7 — Work as an Integral

Work done by a variable force along a path:

$W = int_{x_1}^{x_2} F(x),dx$

For a constant force at angle $\theta$ to displacement: $W = Fd\cos\theta$

Work is a scalar quantity measured in Joules (J).

Worked Example

Find the work done by $F(x) = 3x^2$ from $x = 0$ to $x = 2$ m.

$W = int_0^2 3x^2,dx = x^3\Big|_0^2 = 8$ J ✅

Concept Check 🎯

Work as an Integral 🧮

1. A constant force of 10 N pushes an object 5 m. Work done (J)?

2. $W = int_0^2 3x^2,dx = ?$ J

3. $W = int_0^3 4x,dx = ?$ J

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⚛️ Work-Kinetic Energy Theorem

Part 2 of 7 — Kinetic Energy Theorem

$W_{\text{net}} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

The net work done on an object equals its change in kinetic energy.

Worked Example

A 3 kg object accelerates from 2 m/s to 6 m/s. Find the net work done.

$W = \frac{1}{2}(3)(36) - \frac{1}{2}(3)(4) = 54 - 6 = 48$ J ✅

Concept Check 🎯

Kinetic Energy Theorem 🧮

1. A 3 kg object goes from 2 m/s to 6 m/s. Net work done (J)?

2. $KE = \frac{1}{2}(4)(5^2) = ?$ J

3. A 2 kg object has $KE = 25$ J. What is its speed (m/s)? (Hint: $v = \sqrt{2 \cdot KE/m}$)

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⚛️ Potential Energy Functions

Part 3 of 7 — Potential Energy Functions

Potential energy is related to conservative forces: $F(x) = -\frac{dU}{dx}$

Common potential energies:

• Gravitational: $U = mgh$
• Elastic (spring): $U = \frac{1}{2}kx^2$

A force is conservative if work depends only on endpoints, not path.

Worked Example

Given $U(x) = 5x^2$, find $F(x)$.

$F(x) = -\frac{dU}{dx} = -10x$ ✅

This is a restoring force (like a spring with $k = 10$ N/m).

Concept Check 🎯

Potential Energy Functions 🧮

1. A 4 kg object is at height 5 m. Gravitational PE (J)? ($g = 10$ $m/s^{2}$)

2. A spring ($k = 200$ N/m) is compressed 1 m. Elastic PE (J)?

3. If $U(x) = 5x^2$, what is $F(x)$ at $x = 1$? (Give the numerical value in N.)

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⚛️ Conservation of Energy

Part 4 of 7 — Conservation of Energy

For isolated systems with only conservative forces:

$KE_i + U_i = KE_f + U_f$

$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$

If non-conservative forces (friction) act: $KE_i + U_i + W_{nc} = KE_f + U_f$

Worked Example

A ball is dropped from 20 m. Find its speed at the ground. ($g = 10$ $m/s^{2}$)

$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt{2(10)(20)} = 20$ m/s ✅

Concept Check 🎯

Conservation of Energy 🧮

1. A ball falls from 20 m. Speed at the bottom (m/s)? ($g = 10$ $m/s^{2}$)

2. A ball is launched upward at 20 m/s. What speed (m/s) does it have at height 15 m? ($g = 10$ $m/s^{2}$)

3. A ball is thrown upward at 20 m/s. Maximum height reached (m)? ($g = 10$ $m/s^{2}$, answer as integer. Hint: $h = v^2/(2g)$)

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⚛️ Power

Part 5 of 7 — Power

Power is the rate of doing work:

$P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}$

$P_{avg} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$

Unit: Watt (W) = J/s = $kg\cdot m^{2}/s^{3}$

Worked Example

A motor lifts a 100 kg load 10 m in 5 s. Find the average power. ($g = 10$ $m/s^{2}$)

$P = \frac{W}{t} = \frac{mgh}{t} = \frac{100(10)(10)}{5} = 2000$ W ✅

Concept Check 🎯

Power 🧮

1. A motor lifts 100 kg by 10 m in 5 s. Average power (W)? ($g = 10$ $m/s^{2}$)

2. A force of 50 N moves an object at 10 m/s. Instantaneous power (W)?

3. A 1000 W motor runs for 5 s. How much energy (J) does it deliver? Divide your answer by 5 to give energy per second... wait. 1000 W for 5 s = 5000 J. Hmm. Let me redo: A motor delivers 1000 J in 5 s. What is the power (W)?

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⚛️ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Energy Problem-Solving Strategy

1. Identify the system and its initial/final states
2. Determine if mechanical energy is conserved
3. If friction exists, use $W_{nc} = \Delta KE + \Delta PE$
4. Choose appropriate energy types (KE, gravitational PE, elastic PE)
5. Solve algebraically before substituting numbers

Worked Example

A 2 kg block slides down a 5 m frictionless ramp (30° incline) starting from rest. Find the speed at the bottom.

Height: $h = 5\sin 30° = 2.5$ m

$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2(10)(2.5)} = \sqrt{50} \approx 7.07$ m/s ✅

Concept Check 🎯

Problem-Solving Workshop 🧮

1. A spring ($k = 100$ N/m) is compressed 1 m. How much elastic PE (J) is stored?

2. A 0.5 kg ball is launched by this spring. What is the launch speed (m/s)?

3. A 2 kg block slides down a 5 m ramp against friction ($f_k = 6$ N). Net work done by all forces (J)? ($g = 10$, height = 2.5 m.) $W_{net} = mgh - f_k d = 2(10)(2.5) - 6(5)$.

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⚛️ Review & Applications

Part 7 of 7 — Review & Applications

Summary

• $W = \int F,dx$, $W = Fd\cos\theta$
• Work-KE Theorem: $W_{net} = \Delta KE$
• $F = -dU/dx$ for conservative forces
• Conservation: $E_i = E_f$ (no friction)
• Power: $P = dW/dt = Fv$

Worked Example

A 2 kg ball on a spring ($k = 50$ N/m) is released from $x = 2$ m. Find speed at $x = 0$.

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

$v = x\sqrt{k/m} = 2\sqrt{50/2} = 2(5) = 10$ m/s ✅

Concept Check 🎯

Review & Applications 🧮

1. A 2 kg ball on a spring ($k = 50$ N/m) is released from $x = 2$ m. Speed at $x = 0$ (m/s)?

2. $KE = \frac{1}{2}(2)(3^2) = ?$ J

3. A 1000 W engine runs for 0.5 s. Energy delivered (J)?

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Practice

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