A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the plane by an ideal spring that is aligned with the surface and attached
To a wall above the brick. The spring has a spring constant (force constant) of 120 N/m. By how
Much does the spring stretch with the brick attached?
A) 24 cm
B) 240 cm
C) 14 cm
D) 36 cm
E) 360 cm

Question

Quizplus · Accepted Answer

To find the stretch in the spring, we first calculate the force acting along the incline due to gravity, which is $F = mg\sin(\theta)$, where $m = 3.0 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $\theta = 34^\circ$. This force is balanced by the spring force, $F_{\text{spring}} = kx$, where $k = 120 \, \text{N/m}$ and $x$ is the stretch in the spring. Setting these forces equal gives $mg\sin(\theta) = kx$. Solving for $x$ gives $x = \frac{mg\sin(\theta)}{k}$. Substituting the given values, $x = \frac{(3.0)(9.8)\sin(34^\circ)}{120}$ results in approximately $0.14 \, \text{m}$ or $14 \, \text{cm}$.

A 30-Kg Brick Rests on a Perfectly Smooth Ramp Inclined at at 34°