Lesson 3 · 35 min

Weight, Springs and Friction

Three forces turn up in almost every work–energy problem. The weight's work depends only on the change in height. A spring's work depends only on its starting and final stretch. Friction's work depends on how far the particle slides, so the path matters. Each has a formula worth knowing by heart.

Learning objectives

The work of the weight

Near the Earth's surface the weight is constant: \(\colVg{\mathbf{W}} = -W\,\jhat\), with \(y\) measured upward. As Lesson 2 showed, a constant force does work \(\Fvec\cdot(\rvec_2 - \rvec_1)\), so only the vertical displacement counts:

Work of the weight

\[ \colVg{U_W} = -W\,(y_2 - y_1) = -W\,\Delta y \]

Positive when the particle goes down, negative when it goes up, whatever the path in between. \(W = mg\).

Figure 3.1 A \(5\ \text{kg}\) crate moves from \(A\) to \(B\). Drag the handle \(C\) to bend the path. In a vertical plane, the weight's work stays fixed however long or curved the path gets. On a level floor seen from above (with \(\mu_k = 0.4\)), friction's work grows with the length of the path.

Example 3.1 — Any path down

A \(2\ \text{kg}\) ball is thrown from the top of a \(12\ \text{m}\) tower. It rises, curves over and lands on the ground below. Find the work done by its weight from the throw to the landing.

Show solution
\[ \colVg{U_W} = -W\,\Delta y = -(2)(9.81)(0 - 12) = 235.4\ \text{J} \]

The rise and fall above the tower cancel out: on the way up the weight does negative work, on the way back down to the same height it does the same amount of positive work. Only the net drop of \(12\ \text{m}\) remains.

The work of a spring

A linear spring with stiffness \(k\) pulls or pushes with a force proportional to its deformation \(s\), measured from its unstretched length (stretched or compressed, the magnitude is \(F_s = ks\)). The spring always acts to restore its unstretched length, so on the particle at its end, \(\colVe{F} = -ks\) along the direction of increasing \(s\). Then

\[ U = \int_{s_1}^{s_2} (-ks)\,ds \]

Work of a linear spring on the particle

\[ \colVe{U_s} = -\left(\tfrac12 k s_2^2 - \tfrac12 k s_1^2\right) \]

Negative while the deformation grows, positive while the spring relaxes. Stretched or compressed makes no difference: \(s^2\) is the same either way.

Figure 3.2 The force of a spring on the particle, \(F = -ks\), against the deformation \(s\) (positive stretched, negative compressed). The spring's work from \(s_1\) to \(s_2\) is the signed area under the line. Compare it with the tempting wrong formula \(-\tfrac12 k(s_2 - s_1)^2\), which counts only a triangle.

Example 3.2 — The trapezoid, not the triangle

A spring with \(k = 500\ \text{N/m}\) is stretched from \(s_1 = 0.1\ \text{m}\) to \(s_2 = 0.3\ \text{m}\). Find the work done by the spring on the particle attached to it. Then find its work when the particle returns from \(0.3\) to \(0.1\ \text{m}\).

Show solution
\[ \colVe{U_s} = -\tfrac12(500)\left(0.3^2 - 0.1^2\right) = -250(0.08) = -20\ \text{J} \]

On the way back the end points swap: \(+20\ \text{J}\). The spring gives back exactly what it took.

The tempting \(-\tfrac12(500)(0.3 - 0.1)^2 = -10\ \text{J}\) is wrong by a factor of two here. In Figure 3.2 the work is the whole trapezoid between \(s_1\) and \(s_2\): a rectangle \(ks_1(s_2 - s_1) = 10\ \text{J}\) plus a triangle \(\tfrac12 k(s_2 - s_1)^2 = 10\ \text{J}\). The wrong formula drops the rectangle. It gives the right answer only when the spring starts unstretched, \(s_1 = 0\).

The work of friction

Kinetic friction has magnitude \(\colF{F_f} = \mu_k N\) and always points against the sliding, at \(\theta = 180^\circ\). When \(N\) is constant, over a sliding distance \(d\):

Work of kinetic friction (constant \(N\))

\[ \colF{U_f} = -\mu_k N\,d \]

Always negative on a fixed surface, and \(d\) is the distance slid along the path, not the displacement. A longer path means more negative work.

Find \(N\) from the free-body diagram, perpendicular to the surface, where the particle has no acceleration (on a straight path). It is \(mg\) only on a level surface with no other vertical forces. The table after Figure 3.3 sketches the four cases you will meet most: in each one, add up the forces perpendicular to the surface and set the sum to zero.

Figure 3.3 A \(10\ \text{kg}\) block slides \(2\ \text{m}\) down a rough incline. The weight's arrowhead ends at the center of gravity. The incline acts on the block's bottom face, where they touch: its push \(N\) (black) ends there, pushing out of the surface, and friction \(\colF{F_f}\) (orange) starts there, pointing along the surface against the sliding. Perpendicular to the incline nothing accelerates, so \(N = mg\cos\theta\) and \(\colF{F_f} = \mu_k N\). Steepen the incline: \(N\) and the friction force shrink while the weight's work grows.
The normal force on a straight path, from the free-body diagram. In each sketch \(N\) is drawn to the same scale as the weight (and \(P\) equal to the weight, with \(\alpha = 30^\circ\)), so you can see when \(N\) is smaller or larger than \(mg\).
SituationFree-body diagram\(N\)
Level surface, no other vertical forceWN\(mg\)
Incline at angle \(\theta\)WNθ\(mg\cos\theta\)
Level surface, pulled at \(\alpha\) above the horizontal by \(P\)WNPα\(mg - P\sin\alpha\)
Level surface, pushed at \(\alpha\) below the horizontal by \(P\)WNPα\(mg + P\sin\alpha\)

Example 3.3 — Up a ramp at constant speed

A \(10\ \text{kg}\) crate is pulled \(4\ \text{m}\) up a \(30^\circ\) ramp at constant speed by a rope parallel to the ramp. The coefficient of kinetic friction is \(0.2\). Find the work of each force on the crate, and the net work.

Show solution

Forces. \(W = 98.1\ \text{N}\). Perpendicular to the ramp, \(N = W\cos 30^\circ = 84.96\ \text{N}\), so \(\colF{F_f} = 0.2(84.96) = 16.99\ \text{N}\), down the ramp. At constant speed the rope balances friction and the weight's component along the ramp: \(\colP{P} = W\sin 30^\circ + F_f = 49.05 + 16.99 = 66.04\ \text{N}\).

Work of each force over the 4 m
ForceWork
Rope, \(\colP{P}\)\(+(66.04)(4) = +264.2\ \text{J}\)
Weight, \(\colVg{W}\)\(-W\,\Delta y = -(98.1)(4\sin 30^\circ) = -196.2\ \text{J}\)
Normal force, \(N\)\(0\)
Friction, \(\colF{F_f}\)\(-(16.99)(4) = -67.97\ \text{J}\)
Net work\(0\)

The net work is zero, and so is the change in speed. That is no coincidence: it is the principle of Lesson 4.

Example 3.4 — An angled pull reduces friction

A \(20\ \text{kg}\) crate is pulled \(6\ \text{m}\) across a level floor by a force \(P = 100\ \text{N}\) at \(30^\circ\) above the horizontal. The coefficient of kinetic friction is \(0.3\). Find the work of \(P\), the work of friction and the net work.

Show solution

Vertically, \(N + P\sin 30^\circ - mg = 0\), so \(N = 196.2 - 50 = 146.2\ \text{N}\) and \(\colF{F_f} = 0.3(146.2) = 43.86\ \text{N}\).

\[ \colP{U_P} = (100)\cos 30^\circ\,(6) = 519.6\ \text{J}, \qquad \colF{U_f} = -(43.86)(6) = -263.2\ \text{J} \]

The weight and \(N\) are vertical and do no work, so the net work is \(519.6 - 263.2 = 256.5\ \text{J}\). Lifting part of the crate's weight off the floor cut friction from \(58.9\) to \(43.9\ \text{N}\).

Why the path matters for friction only

The weight is the same force wherever the particle is, so its work depends only on the end points. A spring's force depends only on the deformation, so its work too depends only on the starting and final deformation. Friction always points against the motion, so every bit of sliding, in any direction, takes work away: going out and back again costs twice, it never cancels. Lesson 6 calls forces whose work depends only on the end points conservative.

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Key takeaways