A 0.5 kg block is released by a spring (k = 200 N/m, compressed 10 cm), μ = 0.2. Find the speed at release and the stopping distance — Physics problems | PhysLabo
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A block launched by a compressed spring over a rough floor

Medium 10 grade Mechanics / Energy
Problem statement

A block of mass $m = 0.5$ kg is pressed against a light spring of stiffness $k = 200$ N/m compressed by $x = 10$ cm; the block is not attached to the spring. The spring is released and the block moves along a horizontal floor with a coefficient of friction $\mu = 0.2$ (see the figure). Find the speed of the block at the moment the spring has returned to its natural length and the block leaves it, and the distance $s$ the block then travels before stopping. Take $g = 10$ m/s².

A block m pressed against a spring of stiffness k compressed by x, on a rough horizontal floor
Given
  • $m = 0.5$ kg
  • $k = 200$ N/m
  • $x = 10$ cm $= 0.1$ m
  • $\mu = 0.2$
  • $g = 10$ m/s²
  • $v = ?$
  • $s = ?$
Hints
Hint 1 / 2
The energy of the compressed spring $kx^2/2$ is partly turned into kinetic energy and partly spent on the work of friction over the distance $x$.
Hint 2 / 2
After leaving the spring all the kinetic energy goes into the work of friction: $mv^2/2 = \mu mg\,s$.
Solution

The motion has two stages: the block is accelerated by the spring (friction takes away part of the energy) and then slides by inertia until it stops. The block starts moving because $kx = 20$ N $> \mu mg = 1$ N. The work of friction $F_{fr} = \mu mg$ over a distance $l$ is $\mu mg\,l$.

Energy in two stages: acceleration by the spring and sliding to a stop
Energy in two stages: acceleration by the spring and sliding to a stop

1. Stage 1 (distance $x$): the spring energy goes into kinetic energy and the work of friction:

$$\frac{kx^2}{2} = \frac{mv^2}{2} + \mu mg\,x$$$$v = \sqrt{\frac{kx^2 - 2\mu mg\,x}{m}} = \sqrt{\frac{200 \cdot 0.01 - 2 \cdot 0.2 \cdot 0.5 \cdot 10 \cdot 0.1}{0.5}} = \sqrt{3.6} \approx 1.9 \text{ m/s}$$

2. Stage 2: all the kinetic energy goes into the work of friction:

$$\frac{mv^2}{2} = \mu mg\,s \;\Rightarrow\; s = \frac{v^2}{2\mu g} = \frac{3.6}{2 \cdot 0.2 \cdot 10} = 0.9 \text{ m}$$
Answer

$v \approx 1.9$ m/s; $s = 0.9$ m

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