🎯⭐ INTERACTIVE LESSON

Work and Power

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Work and Power - Complete Interactive Lesson

Part 1: Work as an Integral

⚛️ 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.

Concept Check 🎯

Work as an Integral 🧮

A constant force of 10 N pushes an object 5 m. Work done (J)?
$W = int_0^2 3x^2,dx = ?$ J

Concept Check 🔍

Practice

#	Force	Limits
1	$F = 10$ N constant	0 to 5 m
2	$F(x) = 4x$	0 to 3 m
3	(spring)

Challenge Question 📋

Part 2: Kinetic Energy Theorem

⚛️ Work-Kinetic Energy Theorem

Part 2 of 7 — Kinetic Energy Theorem

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

Part 3: Potential Energy Functions

⚛️ 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:

Part 4: Conservation of Energy

⚛️ Conservation of Energy

Part 4 of 7 — Conservation of Energy

For isolated systems with only conservative forces:

$KE_i + U_i = KE_f + U_f$

Part 5: Power

⚛️ Power

Part 5 of 7 — Power

Power is the rate of doing work:

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

Part 6: Problem-Solving Workshop

⚛️ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Energy Problem-Solving Strategy

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

Part 7: Review & Applications

⚛️ Review & Applications

Part 7 of 7 — Review & Applications

Summary

$W = \int F,dx$ , $W = Fd\cos\theta$
Work-KE Theorem:

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

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

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 🧮

A 3 kg object goes from 2 m/s to 6 m/s. Net work done (J)?
$KE = \frac{1}{2}(4)(5^2) = ?$ J
A 2 kg object has $KE = 25$ J. What is its speed (m/s)? (Hint: $v = \sqrt{2 \cdot KE/m}$ )

Concept Check 🔍

Practice

#	Concept	Key Formula
1	Kinetic energy	$KE = \frac{1}{2}mv^2$
2	Work-energy theorem	$W_{net} = \Delta KE$
3	Stopping distance	$W_{friction} = -\Delta KE$

Challenge Question 📋

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 🧮

A 4 kg object is at height 5 m. Gravitational PE (J)? ( $g = 10$ m/s²)
A spring ( $k = 200$ N/m) is compressed 1 m. Elastic PE (J)?
If $U(x) = 5x^2$ , what is $F(x)$ at $x = 1$ ? (Give the numerical value in N.)

Concept Check 🔍

Practice

#	Potential Energy	Force
1	$U = mgh$	$F = -mg$
2	$U = \frac{1}{2}kx^2$	$F = -kx$
3	$U = ax^3$	$F = -3ax^2$

Challenge Question 📋

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

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/s²)

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

Concept Check 🎯

Conservation of Energy 🧮

A ball falls from 20 m. Speed at the bottom (m/s)? ( $g = 10$ m/s²)
A ball is launched upward at 20 m/s. What speed (m/s) does it have at height 15 m? ( $g = 10$ m/s²)
A ball is thrown upward at 20 m/s. Maximum height reached (m)? ( $g = 10$ m/s², answer as integer. Hint: $h = v^2/(2g)$ )

Concept Check 🔍

Practice

#	Scenario	Equation
1	Dropped object	$mgh = \frac{1}{2}mv^2$
2	Spring launch	$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
3	Friction on ramp	$mgh = \frac{1}{2}mv^2 + f_k d$

Challenge Question 📋

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

Unit: Watt (W) = J/s = kg·m²/s³

Worked Example

A motor lifts a 100 kg load 10 m in 5 s. Find the average power. ( $g = 10$ m/s²)

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

Concept Check 🎯

Power 🧮

A motor lifts 100 kg by 10 m in 5 s. Average power (W)? ( $g = 10$ m/s²)
A force of 50 N moves an object at 10 m/s. Instantaneous power (W)?
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)?

Concept Check 🔍

Practice

#	Concept	Formula
1	Average power	$P = W/t$
2	Instantaneous power	$P = Fv$
3	Horsepower conversion	1 hp ≈ 746 W

Challenge Question 📋

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 🧮

A spring ( $k = 100$ N/m) is compressed 1 m. How much elastic PE (J) is stored?
A 0.5 kg ball is launched by this spring. What is the launch speed (m/s)?
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.) $W_{net} = mgh - f_k d = 2(10)(2.5) - 6(5)$ .

Concept Check 🔍

Practice

#	Problem Type	Key Principle
1	Ramp problems	Energy conservation
2	Spring-block systems	Elastic + kinetic energy
3	Friction on a ramp	Work-energy with $W_{nc}$

Challenge Question 📋

W_{n e t} = Δ K E

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 🧮

A 2 kg ball on a spring ( $k = 50$ N/m) is released from $x = 2$ m. Speed at $x = 0$ (m/s)?
$KE = \frac{1}{2}(2)(3^2) = ?$ J
A 1000 W engine runs for 0.5 s. Energy delivered (J)?

Concept Check 🔍

Practice

#	Topic	Key Formula
1	Work integral	$W = \int F,dx$
2	Energy conservation	$KE_i + PE_i = KE_f + PE_f$
3	Power	$P = Fv$

Challenge Question 📋