A 2.7-kg block is hanging stationary from the end of a vertical spring that is a
ID: 2255202 • Letter: A
Question
A 2.7-kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of the spring/mass system is 2.0 J. What is the elastic potential energy of the system when the 2.7-kg block is replaced by a 5.5-kg block? A 2.7-kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of the spring/mass system is 2.0 J. What is the elastic potential energy of the system when the 2.7-kg block is replaced by a 5.5-kg block?Explanation / Answer
here we know
F=mg=kx
[read this ~ proportional to]
=> x ~ m
& P.E or W = 1/2 *kx^2 => W ~x^2
x~ m => x^2 ~m^2
W~m^2
[W( ) read P.E of]
W(2.7 kg)/W(5.5 kg)= [2.7/5.5]^2
W(5.5 kg) = W(2.7 kg) *(5.5/2.7)^2
=1.3 *(5.5/2.7)^2 =5.39 J
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.