Work for MATH 141
Exam Relevance for MATH 141
Work problems (springs, pumping) appear on most MATH 141 exams. Setting up the integral is the main challenge.
Understanding Work in Physics
In physics, work measures the energy transferred when a force moves an object. The basic formula is:
$$W = F \cdot d$$
But this only works when the force is constant. What if the force changes as the object moves? That's where calculus comes in.
Work with Variable Force
When force varies with position, we integrate:
$$W = \int_a^b F(x) \, dx$$
The idea: Break the motion into tiny displacements $dx$. At each position $x$, the force is $F(x)$, so the tiny bit of work is $F(x) \, dx$. Add them all up!
Problem: A force of $F(x) = x^2$ newtons moves an object from $x = 0$ to $x = 3$ meters. Find the work done.
$$W = \int_0^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9$$
$$\boxed{W = 9 \text{ joules}}$$
Springs and Hooke's Law
Springs follow Hooke's Law: the force needed to stretch or compress a spring is proportional to the displacement from its natural length.
$$F(x) = kx$$
where $k$ is the spring constant and $x$ is the displacement from natural length.
Work to stretch a spring from $x = a$ to $x = b$:
$$W = \int_a^b kx \, dx = \frac{1}{2}k(b^2 - a^2)$$
Problem: A spring has a natural length of 20 cm. A force of 25 N is required to hold it stretched to 30 cm. How much work is done stretching it from 20 cm to 25 cm?
Step 1: Find k
The spring is stretched 10 cm = 0.1 m from natural length when F = 25 N.
$$25 = k(0.1) \implies k = 250 \text{ N/m}$$
Step 2: Calculate work
Stretching from natural length (x = 0) to 5 cm = 0.05 m:
$$W = \int_0^{0.05} 250x \, dx = 250 \cdot \frac{x^2}{2} \Big|_0^{0.05}$$
$$= 125(0.0025 - 0) = 0.3125$$
$$\boxed{W = 0.3125 \text{ joules}}$$
Problem: A spring requires 6 J of work to stretch from its natural length to 10 cm beyond. How much work is needed to stretch it from 10 cm to 15 cm beyond natural length?
Step 1: Find k using the given information
$$6 = \int_0^{0.1} kx \, dx = \frac{k(0.1)^2}{2} = 0.005k$$
$$k = \frac{6}{0.005} = 1200 \text{ N/m}$$
Step 2: Calculate work from 0.1 m to 0.15 m
$$W = \int_{0.1}^{0.15} 1200x \, dx = 600x^2 \Big|_{0.1}^{0.15}$$
$$= 600(0.0225 - 0.01) = 600(0.0125) = 7.5$$
$$\boxed{W = 7.5 \text{ joules}}$$
Pumping Liquids
One of the most common work problems involves pumping liquid out of a tank. The key insight: different layers of liquid must be lifted different distances.
Strategy:
- Set up a coordinate system (usually $y$ = height from bottom)
- Find the volume of a thin slice at height $y$
- Find the weight of that slice: $\text{weight} = \rho g \cdot \text{volume}$
- Find the distance that slice must be lifted
- Integrate!
Problem: A rectangular tank is 4 m long, 2 m wide, and 3 m deep. It's filled with water (density 1000 kg/m³). Find the work to pump all the water over the top edge.
Step 1: Set up coordinates
Let $y$ = height from the bottom, with $y = 0$ at the bottom and $y = 3$ at the top.
Step 2: Volume of a thin slice
A slice at height $y$ with thickness $dy$: $$dV = 4 \cdot 2 \cdot dy = 8 \, dy \text{ m}^3$$
Step 3: Weight of the slice
$$dW_{\text{slice}} = \rho g \cdot dV = 1000 \cdot 9.8 \cdot 8 \, dy = 78400 \, dy \text{ N}$$
Step 4: Distance to lift
A slice at height $y$ must be lifted to height 3, so distance = $3 - y$.
Step 5: Integrate
$$W = \int_0^3 78400(3 - y) \, dy = 78400 \int_0^3 (3 - y) \, dy$$
$$= 78400 \left[3y - \frac{y^2}{2}\right]_0^3 = 78400 \left(9 - \frac{9}{2}\right)$$
$$= 78400 \cdot \frac{9}{2} = 352800$$
$$\boxed{W = 352{,}800 \text{ joules}}$$
Problem: A cylindrical tank of radius 2 m and height 5 m is filled with water. Find the work to pump all water to a height 1 m above the top of the tank.
Step 1: Set up coordinates
Let $y$ = height from bottom. Tank goes from $y = 0$ to $y = 5$.
Step 2: Volume of slice
Each slice is a disk of radius 2: $$dV = \pi (2)^2 \, dy = 4\pi \, dy$$
Step 3: Weight of slice
$$\text{Weight} = 1000 \cdot 9.8 \cdot 4\pi \, dy = 39200\pi \, dy$$
Step 4: Distance to lift
Water must reach height 6 m (1 m above top). Slice at height $y$ travels distance $6 - y$.
Step 5: Integrate
$$W = \int_0^5 39200\pi (6 - y) \, dy = 39200\pi \left[6y - \frac{y^2}{2}\right]_0^5$$
$$= 39200\pi \left(30 - \frac{25}{2}\right) = 39200\pi \cdot \frac{35}{2}$$
$$= 686000\pi \approx 2{,}155{,}133$$
$$\boxed{W = 686{,}000\pi \approx 2{,}155{,}133 \text{ joules}}$$
Problem: A conical tank (vertex down) has radius 3 m at the top and height 6 m. It's full of water. Find the work to pump all water over the top.
Step 1: Set up coordinates
Let $y$ = height from the vertex (bottom), with $y = 0$ at vertex, $y = 6$ at top.
Step 2: Find radius at height y
By similar triangles: $\frac{r}{y} = \frac{3}{6} = \frac{1}{2}$, so $r = \frac{y}{2}$
Step 3: Volume of slice
$$dV = \pi r^2 \, dy = \pi \left(\frac{y}{2}\right)^2 dy = \frac{\pi y^2}{4} \, dy$$
Step 4: Weight of slice
$$\text{Weight} = 1000 \cdot 9.8 \cdot \frac{\pi y^2}{4} \, dy = 2450\pi y^2 \, dy$$
Step 5: Distance to lift
Slice at height $y$ must reach height 6, so distance = $6 - y$.
Step 6: Integrate
$$W = \int_0^6 2450\pi y^2 (6 - y) \, dy = 2450\pi \int_0^6 (6y^2 - y^3) \, dy$$
$$= 2450\pi \left[2y^3 - \frac{y^4}{4}\right]_0^6$$
$$= 2450\pi \left(2(216) - \frac{1296}{4}\right) = 2450\pi (432 - 324)$$
$$= 2450\pi \cdot 108 = 264600\pi$$
$$\boxed{W = 264{,}600\pi \approx 831{,}265 \text{ joules}}$$
Lifting Cables and Chains
When lifting a hanging cable or chain, different parts travel different distances.
Problem: A cable 50 m long weighing 2 kg/m hangs from the top of a building. Find the work to pull the entire cable to the top.
Step 1: Set up coordinates
Let $y$ = distance from top of building. The cable hangs from $y = 0$ to $y = 50$.
Step 2: Weight of a small piece
A piece of length $dy$ at distance $y$ weighs $2 \cdot 9.8 \, dy = 19.6 \, dy$ N.
Step 3: Distance to lift
That piece must be lifted distance $y$ to reach the top.
Step 4: Integrate
$$W = \int_0^{50} 19.6y \, dy = 19.6 \cdot \frac{y^2}{2} \Big|_0^{50}$$
$$= 9.8 \cdot 2500 = 24500$$
$$\boxed{W = 24{,}500 \text{ joules}}$$
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting that x = 0 means natural length for springs
Wrong: Using the total length of the spring as $x$ in Hooke's Law.
Why it's wrong: In $F = kx$, $x$ represents the displacement from natural length, not the total length.
Correct: If a spring has natural length 20 cm and is stretched to 30 cm, then $x = 10$ cm = 0.1 m.
❌ Mistake: Using the wrong distance in pumping problems
Wrong: Using $y$ as the distance to lift when coordinates start at the top.
Why it's wrong: The distance depends on your coordinate setup. If $y = 0$ is at the bottom and the water exits at height $h$, distance = $h - y$.
Correct: Always ask: "How far does THIS slice travel?" Draw a picture!
❌ Mistake: Using constant force for variable force problems
Wrong: $W = F \cdot d$ where $F$ is just the force at one point.
Why it's wrong: If force varies, you can't use a single value. The force at the start might be different from the force at the end.
Correct: Use $W = \int_a^b F(x) \, dx$ whenever force depends on position.
❌ Mistake: Forgetting $g$ in pumping problems
Wrong: Using $\rho \cdot V$ as force instead of $\rho g \cdot V$.
Why it's wrong: Mass isn't force! Weight = mass × $g$ = $\rho V g$.
Correct: Force (weight) = $\rho g \cdot \text{volume}$, where $g \approx 9.8$ m/s².
Work with Variable Force
Work done by a force F(x) that varies with position, moving an object from x = a to x = b.
Variables:
- $W$:
- work done (in joules)
- $F(x)$:
- force as a function of position
- $a, b$:
- starting and ending positions
Hooke's Law
The force required to stretch or compress a spring x units from its natural length. The spring constant k is measured in N/m.
Variables:
- $F(x)$:
- force required (in newtons)
- $k$:
- spring constant (N/m)
- $x$:
- displacement from natural length
Work to Stretch a Spring
Work done stretching a spring from displacement a to displacement b from its natural length.
Variables:
- $W$:
- work done (in joules)
- $k$:
- spring constant (N/m)
- $a$:
- initial displacement from natural length
- $b$:
- final displacement from natural length
Work to Pump Liquid
Work to pump liquid from a tank. Integrate the weight of each slice times the distance it must be lifted.
Variables:
- $W$:
- work done (in joules)
- $ρ$:
- density of liquid (kg/m³)
- $g$:
- acceleration due to gravity (9.8 m/s²)
- $A(y)$:
- cross-sectional area at height y
- $d(y)$:
- distance slice at height y must travel
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