Can someone help me with this Physics problem?

Question

A vertical spring (ignore its mass), whose spring constant is 1000 N/m, is attached to a table and is compressed 0.190 m. 

(a) What speed can it give to a 0.500 kg ball when released? 

(b) How high above its original position (spring compressed) will the ball fly?

Answer 1

(a)

Spring Constant k = 1000 N/m

Compressed distance x is 0.190 m

Mass of the ball attached to 0.5 kg.

According to the law of conservation of energy,

then Total Energy before releasing the bal is equal to Total Energy after releasing the ball.

 

Total Energy before releasing the ball is (1/2)kx²

Total Energy after releasing the ball will have both kinetic energy and potential energy = (1/2)mv² + mgx

(1/2)kx² = (1/2)mv²+ mgx

(1/2) * 1000 * (0.19)² = (1/2)*0.5 * v² + 0.5*9.8*0.190

18.5 = 0.25v² + 0.931

18.5 - 0.931 = 0.25 v²

17.569 = 0.25 v²

v² = 17.569/0.25 = 70.276

v = 8.383

Therefore the speed of the ball when released  is 8.383 m/sec.

Answer 2

(b)

Compressed distance x is 0.190 m

Speed of the ball v = 8.383 m/sec         [from (a)]

If x is the initial compression and h is the height reaches above the point, then conservation of energy implies

Total Energy before releasing the ball is (1/2)kx²

Total Energy after releasing the ball at a height h is mgh

(1/2)kx² = mgh

(1/2)kx² = mgh

h = kx²/2mg

h = (1000*(0.19)²)/(2*0.5 * 9.8)

h = 36.1 / 4.9

h = 3.6836 m

Therefore high above its original position is 3.6836m.