A 260 g block hangs from a spring with spring constant 6.0N/m. At t=0 s the bloc

Question

A 260 g block hangs from a spring with spring constant 6.0N/m. At t=0 s the block is 29 cm below the equilibrium point and moving upward with a speed of 80.0 cm/s.
What is the block's distance from equilibrium at t=2.0 s? A 260 g block hangs from a spring with spring constant 6.0N/m. At t=0 s the block is 29 cm below the equilibrium point and moving upward with a speed of 80.0 cm/s.
What is the block's distance from equilibrium at t=2.0 s?
What is the block's distance from equilibrium at t=2.0 s?

Explanation / Answer

The distance is calculated

As follows

d = vt - y

= 80×2 - 29

= 131 cm

= 1.31 m

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