A2.0 g marble is fired vertically upward using a spring gun. the spring must be compressed 3.1 cm if the marble is to just reach a target 19 m above the marble's position on the compressed spring. (a) what is the change δug in the gravitational potential energy of the marble-earth system during the 19 m ascent? (b) what is the change δus in the elastic potential energy of the spring during its launch of the marble? (c) what is the spring constant of the spring? (a) number enter your answer for part (a) in accordance to the question statement units choose the answer for part (a) from the menu in accordance to the question statement

a) 0.37 J

b) -0.37 J

c) 770.0 N/m

Explanation:

a) The change in the gravitational potential energy during the 19m ascent can be calculated by:

ΔUg = mgd2 - mgd1

Where m is the mass of the marble (2.0 g = 0.002 kg), g is the gravity acceleration (9.8 m/s²), d2 is the final position of the marble (19 m) and d1 is the initial position (3.1 cm = 0.031 m). The distance 0 is the point that was compressed.

ΔUg = mg*(d2 - d1)

ΔUg = 0.002*9.8*(19 - 0.031)

ΔUg = 0.37 J

b) The kinetic energy is 0 because the velocity is equal to 0 at the beginning and final state, so for a conservative system, the change in gravitational potential energy plus the change in elastic potential energy must be 0:

ΔUg + ΔUs = 0

ΔUs = - ΔUg

ΔUs = -0.37 J

c) The change in elastic potential energy is equal to the work done:

ΔUs = w = -(1/2)kx²

Where k is the spring constant, and x is the deformation (3.1 cm = 0.031 m)

-0.37 = -(1/2)*k*(0.031)²

0.000961k = 0.74

k = 770.0 N/m