A block slides down a frictionless 2 m hill and then along a stretch with μ = 0.25. Find the speed at the bottom and the stopping distance — Physics problems | PhysLabo
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Speed at the bottom of a hill and stopping distance

Medium 10 grade Mechanics / Energy
Problem statement

A block slides without friction down a hill of height $h = 2$ m and then, passing smoothly onto a horizontal stretch, slides along it until it stops. The coefficient of friction on the horizontal stretch is $\mu = 0.25$ (see the figure). Find the speed of the block at the foot of the hill and the distance $s$ it travels along the horizontal stretch. Take $g = 10$ m/s².

A frictionless hill of height h and a horizontal rough stretch with friction μ; the block stops at distance s from the foot
Given
  • $h = 2$ m
  • $\mu = 0.25$
  • $g = 10$ m/s²
  • $v = ?$
  • $s = ?$
Hints
Hint 1 / 2
There is no friction on the hill, so mechanical energy is conserved: $mgh = mv^2/2$.
Hint 2 / 2
On the horizontal stretch the work of friction $\mu mg\,s$ equals the lost kinetic energy (or the initial potential energy). The mass cancels.
Solution

There is no friction on the hill, so mechanical energy is conserved, while on the horizontal stretch all the kinetic energy goes into the work of friction. Compare three moments: $A$ — the top, $B$ — the foot, $C$ — the stop.

Energy at points A, B and C: all three quantities are equal
Energy at points A, B and C: all three quantities are equal

1. From $A$ to $B$ (energy is conserved):

$$mgh = \frac{mv^2}{2} \;\Rightarrow\; v = \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 2} = \sqrt{40} \approx 6.3 \text{ m/s}$$

2. From $B$ to $C$ the kinetic energy turns completely into the work of friction $F_{fr} = \mu mg$ over the distance $s$:

$$\frac{mv^2}{2} = \mu mg\,s$$

Or directly from $A$ to $C$: $mgh = \mu mg\,s$ — the mass cancels:

$$s = \frac{h}{\mu} = \frac{2}{0.25} = 8 \text{ m}$$
Answer

$v \approx 6.3$ m/s; $s = 8$ m

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