A spring of spring constant 128 N/m is compressed a distance of 2.0 m from its e

Question

A spring of spring constant 128 N/m is compressed a distance of 2.0 m from its equilibrium

(unstretched, uncompressed) position, and used to project a ball of mass 4.0 kg directly upwards.

Neglect air resistance.

1. [5 pts] What is the potential energy of the spring in its compressed position?

2. [8 pts] To what maximum height above its initial (compressed) position does the ball

reach? (At this height, the ball is no longer in contact with the spring.)

3. [12 pts] Earlier, just when the spring returned to its equilibrium position, as the ball

was moving upwards, how fast was the ball moving?  

The answers are (1):256J (2)6.5m(3)9.4m/s

please show the work

Explanation / Answer

we are given with
k = 128 N / m
x = 2.0 m
m = 4.0 kg
(a)
PE = (1 / 2) k x2 = 0.5*128*2*2 = 256 J
(b)
PE = KE
= (1 / 2) mv2
v = .11.313 m / s
v = ?(2 g h)
h = .6.53 m

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