Work: constant and variable force

The energy section opens with this concept. Work is the energy a force transfers to an object while displacing it. If there is a force but the object does not move, there is no work. If the object moves but the force is perpendicular to the motion, there is again no work. The questions from this section on the exam are usually of four types: work by an angled force, work with the scalar product, the area under a graph and spring work.

First know these four rules:

1. Only the component in the direction of the displacement does work. $W = Fd\cos\theta$.
2. Work is a scalar, it has a sign: positive, negative or zero. It has no direction.
3. If the force varies, the work is the area under the F-x graph: $W = \int F\,dx$.
4. Spring work is $\tfrac12 kx^2$, not $kx \cdot x$. Because the spring force increases as it stretches.

Work by a constant force

A constant force $\vec F$ is applied to an object and the object is displaced by $\vec d$. If the angle between them is $\theta$:

$W = Fd\cos\theta \qquad [W] = \text{N}\cdot\text{m} = \text{J (joule)}$

$F\cos\theta$ is the component of the force in the direction of motion. The perpendicular component $F\sin\theta$ tries to lift or press the object but does no work because there is no motion in that direction.

Positive work gives energy to the object (tries to speed it up), negative work takes energy from the object (tries to slow it down).

PITFALL · work and “getting tired” are not the same thing. If you walk horizontally holding a 20 kg bag, the force your hand exerts on the bag (upward) is perpendicular to the displacement (horizontal). The work in physics is zero. If you push on a wall and cannot move it, the work is zero too.

Net work

If more than one force acts on the object, you calculate the work of each force separately and add them. The same result also comes from the work of the net force:

$W_{net} = \sum W_i = F_{net}\, d\cos\theta$

Worked example 1. A 15.0 kg crate is pulled 12.0 m by a 50.0 N rope that makes an angle of $30.0^\circ$ above the horizontal. $\mu_k = 0.200$. Find the work of each force and the net work.

Work of the rope:  W_F = F d cos30 = 50.0 · 12.0 · 0.866 = 520 J
Normal force:      vertical balance  N + F sin30 - mg = 0
                   N = 15.0·9.80 - 50.0·0.500 = 147 - 25.0 = 122 N
Friction:          f = mu_k N = 0.200 · 122 = 24.4 N
                   W_f = -f d = -24.4 · 12.0 = -293 J
Gravity, N:        perpendicular to the motion → 0 J
Net work:          520 - 293 = 227 J

Pulling upward at an angle makes the object lighter, $N$ gets smaller, and friction is reduced. If you take $N = mg$, you find too much friction.

Work with the scalar product

If the force and the displacement are given in components, there is no need to find the angle. Work is the scalar product (dot product) of the two vectors:

$W = \vec F\cdot\vec d = F_xd_x + F_yd_y + F_zd_z$

This is exactly the dot product of week 2. Unit vector rules: $\hat i\cdot\hat i = 1$, $\hat i\cdot\hat j = 0$.

Worked example 2. $\vec F = (5\hat i + 2\hat j)\ \text{N}$, $\vec d = (3\hat i - 4\hat j)\ \text{m}$. What are the work and the angle?

$W = (5)(3) + (2)(-4) = 15 - 8 = 7\ \text{J}$

$\cos\theta = \frac{W}{|\vec F||\vec d|} = \frac{7}{\sqrt{29}\cdot 5} = 0.260 \;\Rightarrow\; \theta = 74.9^\circ$

PITFALL · negative components. Whoever drops the minus in the $(2)(-4)$ term finds $W = 23$ J. When multiplying, write each component with its sign.

Work by a variable force

If the force varies with position, you cannot use $W = Fd$. Divide the path into very small $\Delta x$ pieces; in each piece the force is almost constant, and add the works. In the limit this becomes an integral:

$W = \int_{x_i}^{x_f} F_x(x)\,dx = \text{the area under the F-x graph}$

It comes in two forms on the exam:

• A formula is given: take the integral. If $F = 2x + 1$, from $x = 1$ to $3$: $W = \left[x^2 + x\right]_1^3 = 12 - 2 = 10\ \text{J}$.
• A graph is given: divide the area into triangle, rectangle and trapezoid pieces. Area below the axis is negative work.

TIP · check the units. If F is in newtons and x in meters in the graph, the area is directly in joules. If x is in centimeters, divide the area by 100.

Spring force and Hooke’s law

If you stretch or compress an ideal spring by $x$ from its equilibrium position, the spring pulls you back. Hooke’s law:

$F_s = -kx$

$k$ is the spring constant, with the unit N/m. A stiff spring has a large $k$. The minus sign says that the force is always toward the equilibrium position: if you stretch it, it pulls back; if you compress it, it pushes.

Since the force increases linearly with $x$, the F-x graph is a line and the area under it is a triangle:

$W_{\text{ext}} = \int_0^x kx'\,dx' = \tfrac12 kx^2$

This is the work done by you to stretch the spring. The work of the spring’s own force is the negative of this (the spring opposes your motion). For a general interval:

$W_{\text{ext}} = \tfrac12 k\left(x_f^2 - x_i^2\right) \qquad W_{\text{spring}} = -\tfrac12 k\left(x_f^2 - x_i^2\right)$

Worked example 3. A spring with $k = 300\ \text{N/m}$ is already stretched 4.00 cm. What work is needed to stretch it to 10.0 cm?

x_i = 0.0400 m, x_f = 0.100 m   (first convert to meters)
W = ½ k (x_f² - x_i²) = ½ · 300 · (0.0100 - 0.00160)
W = 150 · 0.00840 = 1.26 J
The work done by the spring: -1.26 J

PITFALL · $\tfrac12 k(\Delta x)^2$ is wrong. If you write $\tfrac12 k(0.06)^2$ when the spring is already stretched, you assume the spring is being stretched from the start. Square separately, then subtract: $x_f^2 - x_i^2$, not $(x_f - x_i)^2$.

Measuring the spring constant

If a mass $m$ is hung on a spring and the spring stretches by $x$ and stays in equilibrium, the spring force balances the weight: $kx = mg$, $k = mg/x$. This comes up often in lab questions. The extension is given in centimeters; do not forget to convert to meters.

Which method in which case

Mixed exam examples

Worked example 4 (inclined plane). A 3.00 kg block is pulled 5.00 m up a $20.0^\circ$ incline by a 40.0 N force directed up along the incline. $\mu_k = 0.150$. What are the work of each force and the net work?

First the free-body diagram: the pulling force $F$ (up, along the incline), the weight $mg$ (vertically downward), the normal force $N$ (perpendicular to the surface), friction $f_k$ (down, along the incline, because the block is sliding up).

Pull:       W_F = 40.0 · 5.00 = 200 J               (theta = 0)
Gravity:    W_g = -m g d sin20 = -3.00·9.80·5.00·0.342 = -50.3 J
            (rise h = d sin20 = 1.71 m, W_g = -m g h)
Normal:     N = m g cos20 = 27.6 N,  W_N = 0        (perpendicular)
Friction:   W_f = -0.150 · 27.6 · 5.00 = -20.7 J
Net work:   200 - 50.3 - 20.7 = 129 J

You can find the work of gravity in two ways. Either you multiply the component of the weight along the incline ($mg\sin\theta$) by the path, or you multiply the weight by the vertical rise. Both give the same result. This is the first hint of the idea you will see in week 12, “the work of gravity is independent of the path”.

Worked example 5 (a graph that goes below the axis). A force is a constant $+6.00$ N from $x = 0$ to $2.00$ m, then decreases linearly to $-6.00$ N at $x = 4.00$ m and stays at $-6.00$ N until $x = 5.00$ m. What is the total work between 0 and 5.00 m?

0-2 m   rectangle:  6.00 · 2.00          = +12.0 J
2-3 m   triangle (above): ½ · 1.00 · 6.00    =  +3.0 J   (F = 0 at x = 3)
3-4 m   triangle (below): ½ · 1.00 · (-6.00) =  -3.0 J
4-5 m   rectangle:  1.00 · (-6.00)       =  -6.0 J
Total:                                      +6.0 J

It is essential to find the point where the force crosses zero. Whoever adds all the areas as positive finds 24 J.

TIP · the place of this topic on the exam. Work questions are usually the first step of an energy question. Even in a question that ends with “What is the speed?”, you first need to write the work of each force correctly. The work-energy theorem of week 11 is built on top of this calculation.

Common mistakes