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As a ball falls, it speeds up and drops lower — gaining kinetic energy while losing potential energy. Does the total stay fixed?
- Mechanical energy = kinetic energy + potential energy. For a ball dropped from height \(h\): at the top, \(KE = 0\), \(PE = mgh\), so mechanical energy \(= mgh\).
- After falling for time \(t\), the lost potential energy \(\tfrac{1}{2}mg^2t^2\) exactly equals the gained kinetic energy \(\tfrac{1}{2}mg^2t^2\), so the total stays \(mgh\).
- Conservation of mechanical energy: if no other external forces act, the mechanical energy of an object stays constant as it moves under gravity.
In this Activity, we will release a pendulum bob from a marked height and watch whether it swings back up to almost the same level, to test conservation of mechanical energy.
- Set up a simple pendulum. On paper behind it, draw a horizontal line at the level of the resting bob's starting height.
- Take the bob to one side to point P at that line and release it. Watch the extreme points of the first few swings.
- At P the bob has only potential energy (\(mgh\)); at the bottom Q only kinetic energy; at R on the far side only potential energy again. The bob reaches almost the same height — mechanical energy is conserved.
- In real life it slowly stops, because friction at the support and air resistance remove energy.
- Conservation of mechanical energy — if no external forces act, the total mechanical energy (kinetic + potential) of an object remains constant.
- Solving problems directly with Newton's laws can get cumbersome. By keeping track of total mechanical energy, we can often find an object's final speed or position without working through every step of the motion.
What is the speed of a child at the bottom of a slide of height h?
PE at top \(= mgh\) converts to KE at bottom \(= \tfrac{1}{2}mv^2\) (ignoring friction):
\( \tfrac{1}{2}mv^2 = mgh \Rightarrow v = \sqrt{2gh} \).
The speed depends only on the height \(h\) — not on the shape of the slide or the child's mass.
A 10000 kg truck at 72 km h⁻¹ runs onto a 30° sand ramp (sand force 50000 N). Minimum ramp length to stop it? g = 10, and the truck rises 1 m for every 2 m along the ramp.
\(v = 72\ \text{km h}^{-1} = 20\ \text{m s}^{-1}\). Initial KE \(= \tfrac{1}{2}\times 10000 \times 20^2 = 2000000\ \text{J}\).
For ramp length \(d\): height gained \(= d/2\), so PE gained \(= mg(d/2) = 10000\times10\times(d/2) = 50000\,d\). Work by sand \(= -50000\,d\).
Work-energy: \(-50000\,d = (50000\,d) - 2000000 \Rightarrow 2000000 = 100000\,d \Rightarrow d = 20\ \text{m}\).
- Mechanical energy is just one part of a bigger picture. In nature, the total energy of an object or system not acted on by external forces stays constant — energy only changes form.
- 7. Mechanical energy of the ball just before it hits the ground: at the ground \(h=0\) so \(PE=0\), and \(KE = \tfrac{1}{2}mv^2 = mgh\) (since \(v^2 = 2gh\)). So mechanical energy \(= mgh\) — the same as at the top.
- 8. Roller-coaster ball (A, B, C…): at high points PE is large and KE small; at low points KE is large and PE small. Each later peak (C, D, E) is lower because some mechanical energy is lost to friction (and air resistance) along the track.
NCERT Question 3 — When a ball thrown upwards
At the highest point, tick the correct statement(s) about the force, acceleration, kinetic energy and potential energy of the ball.
View the answer →NCERT Question 14 — The potential energy-displacement graph of
A 0.5 kg ball on a frictionless track has v = 0 and PE = 30 J at O (Fig. 7.39). Find its velocity at P, Q and R.
View the answer →NCERT Question 15 — A coconut of mass 1.5
A 1.5 kg coconut falls 10 m onto sand (g = 10). Find its speed before impact, and the depth of the depression if the sand resists with 3000 N.
View the answer →