Asked by Ricky
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?
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?
Answers
Answered by
Anonymous
see your previous post for clues...
Answered by
Anonymous
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)
Answered by
boi
boi what da hell boi
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.