---
title: "Conservative force and pendulum energy"
subject: "Engineering Mechanics"
chapter: 3
key: "semester_3/engineering_mechanics:pendulum-energy-conservation-tension"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/pendulum-energy-conservation-tension
site: BCT Engineering
---

# Conservative force and pendulum energy

Engineering Mechanics · Chapter 3: Kinetics of Particles: Work Energy Principles

- Asked in: year not given, 3+5 marks
- Video: [Conservation of Energy: Free Fall, Springs, and Pendulums](https://www.youtube.com/watch?v=OTK9JrKC6EY) — Professor Dave Explains
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/pendulum-energy-conservation-tension
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-3

## Question

(a) Define conservative force and potential energy. Write expressions for gravitational and elastic potential energy and state the principle of conservation of mechanical energy. (b) A bob of mass 0.5 kg hangs from a light inextensible string 1.2 m long. It is released from rest with the string horizontal. Find the speed of the bob and the tension in the string when the string becomes vertical.

## Short answer

With the lowest point as datum, conservation of mechanical energy gives speed v = sqrt(2gL) = 4.85 m/s. At the vertical position, centripetal force requirement yields tension T = mg + mv²/L = 14.71 N (= 3mg).

## Answer

#### (a) Conservative force and potential energy
A **conservative force** is a force whose work between two points depends only on the end positions and not on the path followed (for example gravity and spring force). Work done around any closed path is zero. Friction is non-conservative.

**Potential energy** $V$ is the energy a body has because of its position or configuration. The work of a conservative force equals the decrease in potential energy: $U_{1\to2} = V_1 - V_2$.

- Gravitational: $V_g = mgh$ (h measured from a chosen datum)
- Elastic (spring of stiffness $k$, extension $x$): $V_e = \tfrac12 kx^2$

**Conservation of mechanical energy:** if only conservative forces do work, the total mechanical energy remains constant:

$$
T_1 + V_1 = T_2 + V_2
$$

#### (b) Pendulum bob
Take the lowest point as datum. Tension does no work (it is perpendicular to the path), so energy is conserved.

```text
   O ____________ (start, v = 0)
   |  \
   |   \ L = 1.2 m
   |    \
   v     o   <- lowest point
```

$$
mgL = \tfrac12 m v^2 \;\Rightarrow\; v = \sqrt{2gL} = \sqrt{2 \times 9.81 \times 1.2} = 4.85\ \text{m/s}
$$

At the lowest point the net upward force provides the centripetal force:

$$
T - mg = \frac{mv^2}{L}
$$

$$
T = 0.5 \times 9.81 + \frac{0.5 \times (4.852)^2}{1.2} = 4.905 + 9.810 = 14.71\ \text{N}
$$

This equals $3mg$, a known result for release from the horizontal.

**Answer: v = 4.85 m/s; T = 14.71 N (= 3mg).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A conservative force is one whose:
   A. Work depends on the path taken
   B. Work between two points is independent of the path (zero around a closed path)
   C. Magnitude is constant
   D. Direction is always vertical
   Answer: B. Gravity and spring forces are conservative; friction is not. (Key point: Conservative force)

2. The gravitational and elastic potential energies are:
   A. mgh and (1/2) k x^2
   B. mg and k x
   C. (1/2) m v^2 and k x
   D. m g h^2 and (1/2) k x
   Answer: A. V_g = mgh and V_e = (1/2)kx^2. (Key point: Potential energies)

3. Bob released from horizontal, L = 1.2 m. Why is the speed at the lowest point sqrt(2gL)?
   A. Tension does work
   B. Tension does no work, so mgL = (1/2) m v^2
   C. Momentum is conserved
   D. Friction cancels gravity
   Answer: B. Tension is perpendicular to the path, so energy is conserved: v = sqrt(2 x 9.81 x 1.2) = 4.85 m/s. (Key point: Speed of bob)

4. The tension at the lowest point is found from:
   A. T = mg
   B. T = m v^2 / L
   C. T - mg = m v^2 / L, giving 3mg = 14.71 N
   D. T + mg = m v^2 / L
   Answer: C. The net upward force provides the centripetal force, so T = mg + mv^2/L = 4.905 + 9.81 = 14.71 N. (Key point: Tension)

## Related in Engineering Mechanics

- [Rod released from horizontal position](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-release-from-horizontal-reaction) — Engineering Mechanics, chapter 6
- [Atwood machine and conservation theorems](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aatwood-machine-lagrange-conservation) — Engineering Mechanics, chapter 9
- [Block lowering a drum, energy method](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Adrum-block-work-energy-acceleration) — Engineering Mechanics, chapter 7
- [Two blocks connected over a pulley](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Ablocks-pulley-friction-acceleration) — Engineering Mechanics, chapter 2
- [Work-energy principle, power and efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Awork-energy-principle-power-efficiency) — Engineering Mechanics, chapter 3
- [Blocks over a massive pulley](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Apulley-blocks-rotational-inertia) — Engineering Mechanics, chapter 6

## Related in other subjects

- [Pendulum bob with given tension](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Apendulum-bob-tension-velocity) — Applied Mechanics (Dynamics), chapter 1
- [Conservation of energy; 2 kg collar on curved rod](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-spring-vertical-2kg) — Applied Mechanics (Dynamics), chapter 2, asked 2×
- [Conservation of energy; 5 kg collar on curved rod](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-spring-vertical-5kg) — Applied Mechanics (Dynamics), chapter 2, asked 2×
- [Pendulum bob velocity and acceleration; momentum](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Apendulum-bob-linear-angular-momentum-tension) — Applied Mechanics, chapter 9, asked 2×