Problem — Block on an inclined plane with a spring (equilibrium and oscillations)

Problem

A block ($m = 5\;\mathrm{kg}$) is attached to a spring ($K = 500\;\mathrm{N/m}$, $\ell_0 = 5\;\mathrm{m}$) on a $30^\circ$ inclined plane $10\;\mathrm{m}$ high. Using the horizontal base as the coordinate $x$, find: (a) the equilibrium point; (b) the turning points for $E_\text{mech} = 2000\;\mathrm{J}$; (c) the maximum speed; (d) the period of small oscillations. Hint: write $E_\text{pot}(x)$ as a parabola in $x$.

Solution

Key hint: write $E_\text{pot}(x)$ as a parabola. The coordinate $x$ is the horizontal projection of the block’s position (measured from the base of the plane). If $d$ is the distance travelled along the plane, then $x = d\cos 30^\circ$, i.e. $d = x/\cos 30^\circ = 1{,}1547\,x$.

• Gravitational potential energy (reference at the base, $h = d\sin 30^\circ = x\tan 30^\circ$):

$E_\text{pot,g} = mg\,x\tan 30^\circ = 5\cdot 9{,}81\cdot 0{,}5774\,x = 28{,}3\,x$

• Elastic potential energy (spring deformation $d - \ell_0 = 1{,}1547\,x - 5$):

$E_\text{pot,m} = \tfrac{1}{2}K\,(1{,}1547\,x - 5)^2 = 250\,(1{,}1547\,x - 5)^2 = 333{,}3\,x^2 - 2886{,}8\,x + 6250$

Adding these gives the parabola of the potential energy:

$E_\text{pot}(x) = 333{,}3\,x^2 - 2858{,}4\,x + 6250 \qquad (a = 333{,}3,\ b = -2858{,}4,\ c = 6250)$

(a) Equilibrium point — vertex of the parabola ($x_\text{eq} = -b/2a$):

$\ev{x_\text{eq} = \frac{2858{,}4}{2\cdot 333{,}3} \approx 4{,}29\;\mathrm{m}}$

The minimum of the potential energy is $E_\text{pot,min} = c - \dfrac{b^2}{4a} \approx 6250 - 6128 \approx 122\;\mathrm{J}$.

(b) Turning points for $E_\text{mech} = 2000\;\mathrm{J}$ — these are the points where $E_\text{pot}(x) = E_\text{mech}$, i.e. $E_\text{kin} = 0$:

$333{,}3\,(x - 4{,}29)^2 + 122 = 2000 \quad\Rightarrow\quad (x - 4{,}29)^2 = \frac{1878}{333{,}3} = 5{,}63$

$x - 4{,}29 = \pm 2{,}37 \quad\Rightarrow\quad \ev{x_1 \approx 1{,}91\;\mathrm{m}, \quad x_2 \approx 6{,}66\;\mathrm{m}}$

(c) Maximum speed — occurs at equilibrium, where kinetic energy is maximum:

$E_\text{kin,max} = E_\text{mech} - E_\text{pot,min} = 2000 - 122 = 1878\;\mathrm{J}$

$\tfrac{1}{2}m\,v_\text{max}^2 = 1878 \quad\Rightarrow\quad v_\text{max} = \sqrt{\frac{2\cdot 1878}{5}} = \sqrt{751{,}2}$

$\ev{v_\text{max} \approx 27{,}4\;\mathrm{m/s}}$

(d) Period of small oscillations. With $E_\text{pot}(x) = a\,x^2 + \dots$ and $\alpha = 30^\circ$ (the angle between the direction of motion, along the plane, and the horizontal coordinate $x$):

$T = 2\pi\sqrt{\frac{m}{2a\cos^2\alpha}} = 2\pi\sqrt{\frac{5}{2\cdot 333{,}3\cdot\cos^2 30^\circ}} = 2\pi\sqrt{\frac{5}{500}} = 2\pi\cdot 0{,}1$

$\ev{T \approx 0{,}628\;\mathrm{s}}$

Collegamenti

Argomenti: Work and energy Concetti: Elastic potential energy · Gravitational potential energy · Conservation of mechanical energy · Elastic force and Hooke’s law · Conservative forces Competenze: Small-oscillation approximation · Reading graphs · Resolving forces on an inclined plane · Symbolic setup Oggetti: Inclined plane · Spring · Block