Problem — Synthesis: inclined plane and spring

Synthesis: inclined plane and spring. A $1\;\mathrm{kg}$ block starts from rest at the top of a $30^\circ$ inclined plane that is $2\;\mathrm{m}$ high, slides down with $\mu_d = 0{,}2$, and at the bottom hits a spring with $k = 500\;\mathrm{N/m}$. Calculate the maximum compression.

Solution

Data: $m = 1\;\mathrm{kg}$, $\theta = 30^\circ$, height $h = 2\;\mathrm{m}$, $\mu_d = 0{,}2$, $k = 500\;\mathrm{N/m}$.

Length of the plane. The block travels down the plane over a stretch

$L = \frac{h}{\sin\theta} = \frac{2}{\sin 30^\circ} = \frac{2}{0{,}5} = 4\;\mathrm{m}$

Energy balance. The initial gravitational potential energy is partly dissipated by friction along the plane, and partly ends up in the spring as elastic energy. The friction force on the plane is $f = \mu_d N = \mu_d\,mg\cos\theta$, and it does dissipative work $Q = f\,L$:

$mgh = Q + \tfrac{1}{2}k x_\text{max}^2, \qquad Q = \mu_d\,mg\cos\theta\;L$

Calculating the terms (using $\cos 30^\circ \approx 0{,}866$):

$mgh = 1\cdot 9{,}81\cdot 2 = 19{,}62\;\mathrm{J}$

$Q = 0{,}2\cdot 1\cdot 9{,}81\cdot 0{,}866\cdot 4 = 6{,}80\;\mathrm{J}$

Energy available for the spring:

$\tfrac{1}{2}k x_\text{max}^2 = mgh - Q = 19{,}62 - 6{,}80 = 12{,}82\;\mathrm{J}$

Maximum compression:

$x_\text{max} = \sqrt{\frac{2(mgh - Q)}{k}} = \sqrt{\frac{2\cdot 12{,}82}{500}} = \sqrt{0{,}0513} \approx 0{,}227\;\mathrm{m}$

$\ev{x_\text{max} \approx 0{,}227\;\mathrm{m} \approx 22{,}7\;\mathrm{cm}}$

Physical reading. Without friction the spring would store all $19{,}62\;\mathrm{J}$ (greater compression); friction removes about $6{,}8\;\mathrm{J}$ of it as heat, reducing the compression. The problem combines three energy reservoirs — gravitational, thermal (friction) and elastic — in a single balance.

Collegamenti

Argomenti: Work and energy Concetti: Conservation of mechanical energy · Elastic potential energy · Thermal energy from friction Competenze: Resolution on the inclined plane Oggetti: Inclined plane · Spring