Worked exercise — apple and spring

Problem

An apple ($m = 0{,}2\;\text{kg}$) falls from a height $h = 3\;\text{m}$ onto a vertical spring ($K = 500\;\text{N/m}$, $\ell_0 = 0{,}4\;\text{m}$). How much does the spring compress?

Solution

Setup. Take the top of the spring as the reference level ($h = 0$). Define two states:

• State A: the apple is at rest at the top ($v = 0$), at height $h = 3\;\text{m}$ above the relaxed spring.
• State B: the apple is again at rest ($v = 0$, point of maximum compression), the spring is compressed by $x$, so the apple is at height $h_B = -x$.

In the absence of friction the total mechanical energy is conserved: $E_{\text{tot}}^A = E_{\text{tot}}^B$.

Energy table (reference: top of the spring):

State	$E_{\text{cin}}$	$E_{\text{pot,g}}$	$E_{\text{pot,m}}$	$E_{\text{tot}}$
A	$0$	$0{,}2\cdot 9{,}81\cdot 3 = 5{,}89$	$0$	$5{,}89$
B	$0$	$-0{,}2\cdot 9{,}81\cdot x = -1{,}96\,x$	$\tfrac{1}{2}\cdot 500\cdot x^2 = 250\,x^2$	$5{,}89$

Conservation equation. Equating the two total energies:

$-1{,}96\,x + 250\,x^2 = 5{,}89$

$250\,x^2 - 1{,}96\,x - 5{,}89 = 0$

Solving (quadratic formula, taking the positive root):

$x = \frac{1{,}96 + \sqrt{1{,}96^2 + 4\cdot 250\cdot 5{,}89}}{2\cdot 250} = \frac{1{,}96 + \sqrt{5893{,}84}}{500} = \frac{1{,}96 + 76{,}77}{500}$

$\ev{x \approx 0{,}158\;\mathrm{m} \approx 16\;\mathrm{cm}}$

The gravitational term $-1{,}96\,x$ is small compared with the elastic one: the spring absorbs almost all the energy of the fall.

Links

Topics: Work and energy Concepts: Gravitational potential energy · Elastic potential energy · Conservation of mechanical energy · Elastic force and Hooke’s law Skills: Using the energy table · Conservation of energy Methods: State-by-state energy table Objects: Spring