A spring gun is able to launch a 7.0 gram marble to a vertical height of 22 mmeasured from the compressed point of the marble). If the spring iscompressed 8.0 cm from its relaxed state, what will be the spring constant

Spring constant = 472N/m

Explanation:

The change in  gravitational potential energy by the spring is given as = mgh

where m= 7.0 g = 7 X 10 -3kg

g= 9.8m/s

h= 22m

Gravitational potential energy= mgh

= 7.0 x  10^-3 X 9.8  x 22 = 1.5092 J

Remember that  change in  gravitational potential energy by the spring =elastic potential energy

Therefore,  Potential energy P. E = 1/2 K x²

where K= CONSTANT

x= 8.0

2 X 1.5092 J / (8.0 X 10^-2)²= 471.625 ROUNDED TO 472 N/m

490 n

f=125n a=2.5m/s2