A block of mass 0.14 kg is placed on a verti- cal spring of constant 5723 N/m an

Question

A block of mass 0.14 kg is placed on a verti-
cal spring of constant 5723 N/m and pushed
downward, compressing the spring 0.08 m.
After the block is released it leaves the spring
and continues to travel upward.
The acceleration of gravity is 9.8 m/s2 .
What height above the point of release will
the block reach if air resistance is negligible?
Answer in units of m

Explanation / Answer

By PE(spring) = PE(gained by block) =>1/2kx^2 = mgh =>h = kx^2/2mg =>h = [5723 x (0.080)^2]/[2 x 0.14 x 9.8] =>h = 13.35m Thus the net height above the point of release(H) = h - x =>H = 13.35 - 0.08 = 13.26m

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