An elastic chord is 65cm long when a weight of 75N hangs from it, but is 85cm long when a weight of 180N hangs from it. What is the "spring" constant 'k' of this elastic chord?
The answer in the book is 5.3 x 10^2 N/m but I don't know why!
2007-01-29 19:50:54 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The spring constant of a spring (or elastic cord, or any other body that you are told obeys Hooke's law) is k, such that F = kx, where F is the force in the spring and x is the extension of the spring. When you're just told that the cord is 65 cm long when 75 N hang from it, you can't calculate k because you don't know the original length of the cord. But with the second set of numbers, you can see that the additional weight is (180 N) - (75 N) = 105 N, and the additional extension is (85 cm) - (65 cm) = 20 cm = 0.2 m. Now we can say that F = kx ==> k = F/x = (105 N) / (20 cm) = (105 N) / (0.2 m) = 525 N/m = 5.3 x 10^2 N/m.
2007-01-29 19:57:12 · answer #1 · answered by DavidK93 7 · 2⤊ 0⤋
The exceess force applied is 180 - 75 = 105N
This excess force has increased the length by (85 - 65) = 20cm
= 0.2 m
k = force per unit displacement.
k = 105 / 0.2 = 525 N/m. = 5.25 x 10^2 N/m.
2007-01-29 21:56:55 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
check up on the formula, i've forgotten those, so i can't really help you, only tell you how to proceed-
take the original length of string as a variable & develop 2 equations of the spring constant from the data.
now solve, and you should get the answer
2007-01-29 19:56:04 · answer #3 · answered by s_kundu88 3 · 0⤊ 2⤋