Q.A spring of force constant $k = 400\ \text{N/m}$ is compressed by $0.2\ \text{m}$ and used to launch a block of mass $1\ \text{kg}$ up a frictionless incline of angle $30^\circ$ to the horizontal. Find the distance the block travels up the incline before it momentarily comes to rest (take $g = 9.8\ \text{m/s}^2$).

Given: spring constant $k=400\ \text{N/m}$, compression $x=0.2\ \text{m}$, block mass $m=1\ \text{kg}$, incline angle $\theta=30^\circ$, frictionless, $g=9.8\ \text{m/s}^2$.

Energy stored in the compressed spring: $U_{\text{spring}} = \frac{1}{2}kx^2 = \frac{1}{2}(400)(0.2)^2 = \frac{1}{2}(400)(0.04) = 8\ \text{J}$

On a frictionless incline, this entire stored energy converts into gravitational PE at the point where the block momentarily comes to rest, a distance $d$ up the incline (vertical height risen $= d\sin\theta$): $U_{\text{spring}} = mgd\sin\theta$ $8 = (1)(9.8)(d)(\sin30^\circ) = (1)(9.8)(d)(0.5) = 4.9d$ $d = \frac{8}{4.9} \approx 1.63\ \text{m}$

Compute the spring's stored elastic PE using $\tfrac12kx^2$, equate it to the block's gravitational PE gain $mgd\sin\theta$ at the highest point on the incline, and solve for $d$.

• Equating the spring PE to $mgd$ directly, forgetting the $\sin\theta$ factor needed to convert distance along the incline into vertical height risen.
• Using $\cos\theta$ instead of $\sin\theta$ for the height component.