Asked by Arianne
An object mass 1,448 g is attached to a spring with spring constant k = 815 N/m. The object is then displaced by a distance d from the equilibrium position and is released. Shortly after, it passes with speed v = 6.4 m/s through the equilibrium position. By what distance d was the object initially displaced?
i am extremely confused on this question..any help is greatly appreciated!!!
i am extremely confused on this question..any help is greatly appreciated!!!
Answers
Answered by
bobpursley
The initial PE was 1/2 kx^2
the equilibrium KE is equal to that
so, 1/2 m 6.4^2=1/2 815*d
put in 1.448kg for m, and solve for d.
the equilibrium KE is equal to that
so, 1/2 m 6.4^2=1/2 815*d
put in 1.448kg for m, and solve for d.
Answered by
Karina
use v=sqrt 2* Etotal/m
solve for Etotal
then use Bernoulli's equation and solve for k
solve for Etotal
then use Bernoulli's equation and solve for k
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