Work and energy: a slide, a rough floor, and where the energy goes

Energy conservation

How fast is a block at the bottom of a slide, and how far does it get on a rough floor?

Move a block down a smooth curved slide and across a rough floor. The bars show potential energy turning into kinetic energy and then into heat; the total never changes, so v = √(2gh) at the bottom and the block stops after h/μ.

Drag the ringed points on the board.

High school
Kinetic and potential energy
Work–energy theorem
Conservation of mechanical energy
Work done by friction

h

3.0

m

1.0

\mu

smooth

0.30

x

-3.00

mgh = \tfrac12 mv^2 + \mu mg\,s

$E_p = mgy$: 7.35
$E_k = \tfrac12 mv^2$: 22.05
$Q = \mu mg\,s$: 0.00
$E_p + E_k + Q$: 29.40
$v$: 6.64

Blockmoving

Guided tour

1Halfway down a smooth slide
A $1\,\text{kg}$ block starts from rest $3\,\text{m}$ up, with $E=mgh=29.4\,\text{J}$ . Partway down it is $0.75\,\text{m}$ high: $E_p=7.35\,\text{J}$ is left as potential energy and the other $22.05\,\text{J}$ has become kinetic, so $v\approx6.64\,\text{m/s}$ .
2At the bottom
At the bottom all $29.4\,\text{J}$ is kinetic: $\tfrac12mv^2=mgh$ gives $v=\sqrt{2gh}\approx7.67\,\text{m/s}$ . The shape of the slide and the mass of the block do not enter; only the drop does.
3Onto a rough floor
Now the floor has $\mu=0.3$ . After $5\,\text{m}$ friction has done $\mu mg\,s=14.7\,\text{J}$ of negative work, turned into heat. Half the energy is gone and $v\approx5.42\,\text{m/s}$ . The bars still add up to $29.4\,\text{J}$ .
4Where it stops
The work–energy theorem from start to stop: $mgh-\mu mg\,d=0$ , so $d=\dfrac h\mu=10\,\text{m}$ . Double the mass to $2\,\text{kg}$ and all $58.8\,\text{J}$ still turns to heat at exactly the same spot.

Explanation

1
The work–energy theorem
The net work done on a body equals the change in its kinetic energy:
$W_{\text{net}}=\tfrac12mv_2^2-\tfrac12mv_1^2$
It only needs the start and the end, not what happens in between, which is why it handles curved slides that Newton's laws would make hard. Gravity does $mgh$ of work on the way down; friction does $-\mu mg\,s$ on the floor.
2
Mechanical energy is conserved, until friction
When only gravity does work, potential plus kinetic energy stays fixed:
$mgh_1+\tfrac12mv_1^2=mgh_2+\tfrac12mv_2^2$
The normal force does no work because it is always perpendicular to the motion. Friction does, and the energy it removes reappears as heat $Q=\mu mg\,s$ . Count the heat and the total is still conserved.
3
Why the mass keeps cancelling
Every term here has an $m$ : $mgh$ , $\tfrac12mv^2$ , $\mu mg\,s$ . Divide it out and the speed at the bottom, $v=\sqrt{2gh}$ , and the stopping distance, $d=h/\mu$ , are the same for a feather-light block and a heavy one, as long as air resistance is ignored.

Check yourself

Q01

A child slides from rest down a smooth slide $5\,\text{m}$ high. Speed at the bottom ( $g=9.8\,\text{m/s}^2$ )?

Q02

A block reaches a rough floor ( $\mu=0.25$ ) at $5\,\text{m/s}$ . How far does it slide?

Q03

On the curved slide, the normal force does

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