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A block of mass 0.240 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.096 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)

User Vikiiii
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1 Answer

1 vote

Answer:

10.19 m

Step-by-step explanation:

Energy is conserved, so elastic energy stored in spring = gravitational energy of block.

1/2 kx² = mgh

h = kx² / (2mg)

h = (5200 N/m) (0.096 m)² / (2 × 0.240 kg × 9.8 m/s²)

h = 10.19 m

User Veniece
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