The Energy Stored on the Shelf
A $2\ \text{kg}$ ball sits on a shelf. Lifting it there cost work, and that work is now stored — release the ball and the store empties back into motion. The two stores of this lesson are gravitational potential energy $mgh$ (bought by height) and kinetic energy $\dfrac{1}{2}mv^{2}$ (carried by speed). Raise the shelf and watch what the climb buys.
stored energy $mgh = 2 \times 9.8 \times h$ = 58.8 J · speed on landing $v = \sqrt{2gh}$ = 7.67 m s⁻¹
The stored energy is $mgh$; if the ball falls freely, that same energy reappears as $\dfrac{1}{2}mv^{2}$. Notice the speed on landing does not depend on the mass.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. A crate is pushed 5 m across a floor by a force of 30 N acting along the motion. The work done is:
R2. A body moves in a straight line at constant speed. The work done by the resultant force is:
R3. A crate of mass 4 kg is lifted onto a shelf 2 m above the floor. The work done against gravity is:
A body of mass $m$ moving with speed $v$ carries kinetic energy
$E_{k} = \dfrac{1}{2}mv^{2}$
| Change | Effect on kinetic energy |
|---|---|
| speed doubles | kinetic energy is multiplied by 4 |
| mass doubles | kinetic energy doubles |
Mini-example: a ball of mass $0.2\ \text{kg}$ at $15\ \text{m s}^{-1}$ carries $\dfrac{1}{2} \times 0.2 \times 15^{2} = 22.5\ \text{J}$. The speed is squared, so speed matters far more than mass.
Raising mass $m$ through a height difference $\Delta h$ stores
$E_{p} = mg\Delta h$
measured from any chosen level — only changes in height matter. Rising stores energy; falling releases it, and the path taken makes no difference.
Mini-example: a gymnast of mass $50\ \text{kg}$ lowered $3.5\ \text{m}$ releases $50 \times 9.8 \times 3.5 = 1715\ \text{J}$ toward her kinetic energy store.
On a smooth slope, $mgh = \dfrac{1}{2}mv^{2}$, so $v = \sqrt{2gh}$ — the mass cancels, and a heavy and a light body released together arrive with the same speed.
$v = \sqrt{2 \times 9.8 \times 5} = 9.9\ \text{m s}^{-1}$, whatever the mass. With resistance present, the speed on arrival is smaller — the missing energy has gone into heat, which the next lessons will audit.
Groups of 4: slide the block down the ramp with the position slider and watch the three bars. On the smooth ramp the GPE bar empties straight into the KE bar; on the rough ramp a heat bar begins to claim its share. Run both ramps, describe what the bars conserve, then answer the three graded challenges together.
Work alone. Four claims about kinetic and potential energy are printed. For each, decide "It is correct" or "There is an error" and then read the explanation. Two of the four are sound — find them without guessing.
A girl of mass $45\ \text{kg}$ slides down a smooth water chute of length $6\ \text{m}$ inclined at $40^{\circ}$ to the horizontal. (i) Find (a) the decrease in her potential energy (b) her speed at the bottom. (ii) How are the answers to part (i) affected if the slide is not smooth?
The drop is the vertical fall down the chute: $6\sin 40^{\circ}$. On a smooth chute every joule of potential energy released becomes kinetic energy, so write $mg\Delta h = \dfrac{1}{2}mv^{2}$ — and notice what cancels.
Vertical drop $= 6\sin 40^{\circ} = 3.857\ \text{m}$, so $mg\Delta h = 45 \times 9.8 \times 3.857 = 1701\ \text{J} \approx 1700\ \text{J}$ released.
Smooth slide: $\dfrac{1}{2} \times 45 \times v^{2} = 1701$, so $v^{2} = 75.6$ and $v = 8.69\ \text{m s}^{-1} \approx 8.7\ \text{m s}^{-1}$.
The decrease in potential energy is unchanged — it depends only on the height. But friction removes energy along the way, so the kinetic energy gained at the bottom is less and the speed is smaller than $8.7\ \text{m s}^{-1}$.
A gymnast of mass $50\ \text{kg}$ swings on a rope of length $10\ \text{m}$. Initially the rope makes $50^{\circ}$ with the vertical. (i) Find the decrease in her potential energy when the rope has reached the vertical. (ii) Find her kinetic energy and hence her speed when the rope is vertical, assuming air resistance may be neglected. (iii) What angle will the rope make with the vertical when she is next temporarily at rest? (iv) Explain why the tension in the rope does no work.
The fall at the lowest point is $10(1 - \cos 50^{\circ})$ — draw the rope at the start and at the vertical and compare heights. Parts (ii) and (iii) are pure energy conservation; for part (iv), ask what direction the gymnast is moving compared with the rope.
At $50^{\circ}$ the gymnast is $10\cos 50^{\circ} = 6.428\ \text{m}$ below the pivot; at the vertical she is $10\ \text{m}$ below. Drop $= 10 - 6.428 = 3.572\ \text{m}$, so the potential energy released is $50 \times 9.8 \times 3.572 = 1750\ \text{J}$.
All of it becomes kinetic energy: $E_{k} = 1750\ \text{J}$. Then $\dfrac{1}{2} \times 50 \times v^{2} = 1750$, giving $v^{2} = 70.0$ and $v = 8.37\ \text{m s}^{-1}$.
No energy is lost, so she rises back to the same height: the rope makes $50^{\circ}$ with the vertical again. The tension is always along the rope while she moves perpendicular to it, so the tension does no work.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A 4 kg object moves at 3 m/s. Its kinetic energy is:
Q2. A 5 kg box is placed on a shelf 2 m above the ground. Its gravitational potential energy relative to the ground is:
Q3. A 0.5 kg ball falls from 4 m to 1 m above a table. Its loss in potential energy is:
Q4. A block slides down a smooth slope of height 5 m from rest. Its speed at the bottom is about:
Work each on paper, then reveal the solution to check your working.
Find the kinetic energy of each of the following. (i) An ice skater of mass $50\ \text{kg}$ travelling with speed $10\ \text{m s}^{-1}$. (ii) An elephant of mass $5\ \text{tonnes}$ moving at $4\ \text{m s}^{-1}$. (iii) A train of mass $7000\ \text{tonnes}$ travelling at $40\ \text{m s}^{-1}$.
Reveal Solution
(i) $\dfrac{1}{2} \times 50 \times 10^{2} = 2500\ \text{J}$. (ii) $\dfrac{1}{2} \times 5000 \times 4^{2} = 40\,000\ \text{J}$. (iii) $\dfrac{1}{2} \times 7\,000\,000 \times 40^{2} = 5.6 \times 10^{9}\ \text{J}$ — five and a half billion joules.
A bricklayer carries a hod of bricks of mass $25\ \text{kg}$ up a ladder of length $10\ \text{m}$ inclined at $60^{\circ}$ to the horizontal. (i) Calculate the increase in the gravitational potential energy of the bricks. (ii) If instead he had raised the bricks vertically to the same height, would the increase in potential energy be (a) less, (b) the same, or (c) more than in part (i)?
Reveal Solution
(i) Vertical rise $= 10\sin 60^{\circ} = 8.660\ \text{m}$, so $mgh = 25 \times 9.8 \times 8.660 = 2122\ \text{J} \approx 2120\ \text{J}$. (ii) (b) the same — potential energy depends only on the height gained, not on the path.
Calculate the change in gravitational potential energy in each case and state whether it is an increase or a decrease. (i) A $2\ \text{kg}$ object moves from $2.6\ \text{m}$ above the base level to $1.2\ \text{m}$ above it. (ii) A $4\ \text{kg}$ object moves from $0.8\ \text{m}$ above the base level to $2.2\ \text{m}$ above it. (iii) A $0.6\ \text{kg}$ object moves from $1\ \text{m}$ below the upper platform to $3\ \text{m}$ below it.
Reveal Solution
(i) $2 \times 9.8 \times (1.2 - 2.6) = -27.4\ \text{J}$ — a decrease. (ii) $4 \times 9.8 \times (2.2 - 0.8) = +54.9\ \text{J}$ — an increase. (iii) The object drops 2 m: $0.6 \times 9.8 \times (-2) = -11.8\ \text{J}$ — a decrease.
$A$ and $B$ are two points on the vertical cross-section of a surface, and $A$ is $5\ \text{m}$ higher than $B$. A particle of mass $0.35\ \text{kg}$ passes through $A$ with speed $7\ \text{m s}^{-1}$, moving on the surface towards $B$. (i) Assuming that there is no resistance to motion, find the speed with which the particle reaches $B$. (ii) Assuming instead that there is a resistance to motion, and that the particle reaches $B$ with speed $11\ \text{m s}^{-1}$, find the work done against this resistance as the particle moves from $A$ to $B$.
Reveal Solution
(i) $\dfrac{1}{2} \times 0.35 \times v^{2} = \dfrac{1}{2} \times 0.35 \times 7^{2} + 0.35 \times 9.8 \times 5$, so $v^{2} = 49 + 98 = 147$ and $v = 12.1\ \text{m s}^{-1}$. (ii) Energy at $A$: $\dfrac{1}{2} \times 0.35 \times 49 = 8.575\ \text{J}$ of kinetic plus $17.15\ \text{J}$ of potential; at $B$ the kinetic energy is $\dfrac{1}{2} \times 0.35 \times 121 = 21.175\ \text{J}$. Work against resistance $= 8.575 + 17.15 - 21.175 = 4.55\ \text{J}$.
A bullet of mass $20\ \text{g}$ had penetrated $16\ \text{cm}$ into a wooden post. Its speed before impact is known to be $80\ \text{m s}^{-1}$. (i) Find the kinetic energy of the bullet before it entered the post. (ii) Write down the work done in stopping the bullet. (iii) Calculate the resistive force on the bullet, assuming it to be constant. A second bullet of the same mass and shape penetrates $20\ \text{cm}$. (iv) Estimate the speed of this bullet before it hit the post.
Reveal Solution
(i) $\dfrac{1}{2} \times 0.02 \times 80^{2} = 64\ \text{J}$. (ii) All of it must be removed: $64\ \text{J}$. (iii) $F = \dfrac{64}{0.16} = 400\ \text{N}$. (iv) Same force over $0.20\ \text{m}$ removes $400 \times 0.20 = 80\ \text{J}$, so $\dfrac{1}{2} \times 0.02 \times v^{2} = 80$, $v^{2} = 8000$ and $v = 89\ \text{m s}^{-1}$ (2 s.f.).
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