An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a distance of 3.0 cm below the starting point. if instead of having been lowered slowly the object was dropped from rest, how far then would it then stretch the spring at maximum elongation?

x = 0.06 m =  6 cm

Explanation:

extension (x) = 3 cm = 0.03 m

force = mg = kx

k (spring constant) = mg / x = mg / 0.03

loss in potential energy = gain in potential energy

mgx = o.5k

mg = 0.5kx

x = mg / 0.5k

recall that k = mg / 0.03

therefore

x =

x = mg x

x =

x = 0.06 m =  6 cm

true is the correct answer

hope i

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat