A 0.50kg mass is suspended on a spring that stretches 3.0cm.
a. What is the spring constant?
b. What added mass would stretch the spring an additional 2.0cm?
c. What is the change in potential energy when the mass is added?
3 answers
see your previous post for clues...
weight = m g = 0.50 * 9.8 = 4.9 N
k = 4.9 Newtons / 0.030 meter
b.) .050 meters total x
m g = (4.9 / 0.03) 0.05
so
m = .5/.03*.05 = 0.83 kg total
0.83 - 0.50 = 0.33 Kg additional
c) U = (1/2) k x^2
change = (1/2)(4.9/.03) (.05^2-.03^2)
k = 4.9 Newtons / 0.030 meter
b.) .050 meters total x
m g = (4.9 / 0.03) 0.05
so
m = .5/.03*.05 = 0.83 kg total
0.83 - 0.50 = 0.33 Kg additional
c) U = (1/2) k x^2
change = (1/2)(4.9/.03) (.05^2-.03^2)
boi what da hell boi
3 answers