Serious help with Physics?

Question

A vertical spring (ignore its mass), whose spring constant is 930 N/m, is attached to a table and is compressed 0.160 m.
(a) What speed can it give to a 0.500 kg ball when released?
m/s
(b) How high above its original position (spring compressed) will the ball fly?

Answer 1

As you well know that
Pe= (1/2) k x^2

a) Ke= Pe(s) - Pe(g)
(1/2)mV^2 = (1/2) k x^2 - mgx

V= sqrt( [ k x^2 - 2mgx]/m)
V=sqrt([930 x (0.160)^2 - 2 x 0.500x9.81 x 0.160]/ 0.500)
b) V=6.67m/s

Ke=Pe
(1/2)mV^2=mgh
h= (1/2g)V^2
h=[(1/ (2x 9.81)] (6.67)^2
h=2.27 m

Answer 2

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spring constant = 930 N/m

compression x=0.160 m

potential energy stored in spring=(1/2)kx^2

potential energy stored in spring=(1/2)930*0.16=74.4 J

kinetic energy of mass = (1/2)mv^2

kinetic energy of mass = (1/2)0.500*v^2 = 0.250v^2

kinetic energy of mass=potential energy in spring

0.25v^2 =74.4

v^2=297.6

v =17.25 m/s

(a) speed given to the ball = 17.25 m/s
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height attained h=v^2/2g

h=297.6 /2*9.8= 15.18 m

height above compressed spring=15.18 +0.16=15.34 m

the ball rises 15.34 m above compressed spring
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