Question

A block of mass $m=15\text{kg}$ is attached to a spring of stiffness $K=100\text{N/m}$. The block descends a plane inclined at an angle $\alpha ={30}^{\circ }$ with horizontal. Assuming there is no friction, determine the acceleration of the block when the spring has stretched by a length $x=0.05\text{m}$. (Acceleration due to gravity is $10{\text{m/s}}^2$)

Answer 1

Mass of the block, $m=15kg$
Stiffness of spring, $K=100\text{N/m}$
Elongation, $x=0.05\text{m}$
FBD diagram

Using Newton's second law,
$ma=mg\text{sin}{30}^{\circ }-kx\Rightarrow 15a=150\times \frac{1}{2}-100\left(0.05\right)\Rightarrow 15a=75-5$
$a=\frac{70}{15}=4.66{\text{m/s}}^2$