Asked by Austin Viens-DeRuisseau
: A 0.98-kg block is held in place against the spring by a 49-N horizontal external force (see the figure). The external force is removed, and the block is projected with a velocity v1 = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 0.32. The velocity of the block is v3 = 1.4 m/s at C. The block moves on to D, where it stops.what is the spring constant of the spring
Answers
Answered by
R_scott
k x = 49 N
1/2 k x^2 = 1/2 m v^2 ... k x^2 = m v^2 = .98 kg * 1.44 m^2/s^2 = 1.41 N⋅m
x = (k x^2) / (k x) = (1.41 N⋅m) / 49 N = .029 m
k = k x / x = 49 n / .029 m
1/2 k x^2 = 1/2 m v^2 ... k x^2 = m v^2 = .98 kg * 1.44 m^2/s^2 = 1.41 N⋅m
x = (k x^2) / (k x) = (1.41 N⋅m) / 49 N = .029 m
k = k x / x = 49 n / .029 m
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.