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

ID: 2254364 • Letter: A

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?

explaining please !!!

Potential energy of the spring in its compressed position is 0.5* k* x^2 =256 J

When the ball reaches its maximum height initial energy of the system = final energy of the system. Final PE= m* g* h

therefore height = 6.53 m

when the spring reached its equilibrium position Energy of the system = Initial PE of the system.

The whole energy is stored as KE of the ball 0.5 * m* v^2 = 256

V=11.31 m/sec

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