Spring on an inclined plane

Problem

A block of $3\;\text{kg}$ on a $33^\circ$ plane is connected to a spring ($K = 100\;\text{N/m}$, natural length $\ell_0 = 1\;\text{m}$, current length $\ell = 0{,}5\;\text{m}$) anchored at the top of the plane. Find the acceleration.

Solution

Data. $m = 3\;\text{kg}$, $\theta = 33^\circ$, $K = 100\;\text{N/m}$, $\ell_0 = 1\;\text{m}$, $\ell = 0{,}5\;\text{m}$.

Elastic force. The spring is compressed ($\ell < \ell_0$): by Hooke’s law it tends to lengthen, so it pushes the block away from the anchor point, i.e. down the plane. $\abs{\vec{F}_m} = K\abs{\ell - \ell_0} = 100 \cdot 0{,}5 = 50\;\text{N}$

Axes. Axis along the plane, positive pointing up. Acting on the block are the weight (component $-mg\sin\theta$), the normal reaction $\vec{R}$ and the elastic force ($-50\;\text{N}$, pointing down).

Newton’s second law along the plane. $-mg\sin33^\circ - 50 = m\,a \quad\Rightarrow\quad -29{,}43 \cdot 0{,}545 - 50 = 3a \quad\Rightarrow\quad -16 - 50 = 3a$ $a = \frac{-66}{3} = -22\;\text{m/s}^2$ The negative sign confirms the block accelerates downhill: weight and compressed spring push in the same direction.

Perpendicular to the plane (constraint reaction): $R = mg\cos33^\circ = 29{,}43 \cdot 0{,}839 = 24{,}7\;\text{N}$

Result. $\ev{a = -22\;\text{m/s}^2 \quad (\text{downhill}), \qquad R = 24{,}7\;\text{N}}$

Links

Topics: Dinamica Concepts: Seconda legge di Newton · Forza elastica e legge di Hooke · Forza peso · Forza normale Skills: Scomposizione sul piano inclinato · Diagramma di corpo libero Objects: Piano inclinato · Molla · Blocco