A verically mounted spring is compressed by .140 m relative to its unstrained length. a mass is placed against the spring at point A. when the spring is released the mass is launced in the air. how fast is the mass moving when it passes a point b that is .880, higher than point a

8.306 m/s

Explanation:

Given

m = mass of the object = 0.200 kg

k = spring constant = 880 N/m

x = length through which the spring is compressed = 0.140 m

h = 0.880 m higher than the original level of the mass

Let the velocity of the mass at the required height be v

Using the law of conservation of Energy

The elasto-potential energy stored in the spring is equal to the sum of work done to raise the mass to any height plus the change in kinetic energy of the mass.

(Elasto-potential Energy stored in the spring) = (Workdone in reaching the height) + (Change in Kinetic energy)

Elasto-potential Energy stored in the spring = ½kx² = (1/2)(880)(0.140²) = 8.624 J

Workdone in reaching 0.880 m above the original position = mgh = (0.2)(9.8)(0.88) = 1.7248 J

Change in kinetic energy = (Final kinetic energy) - (Initial kinetic energy)

Initial kinetic energy = 0 J (since the mass was initially at rest)

Final kinetic energy = ½mv² = (1/2)(0.2)v² = 0.1v²

8.624 = 1.7248 + 0.1v²

0.1v² = 8.624 - 1.7248 = 6.8992

v² = 68.992

v = 8.306 m/s

Hope this Helps!!!

d.7.5cm is your answer