Hooke\'s law describes a certain light spring of unstretchedlength 36.0 cm. When
ID: 1740402 • Letter: H
Question
Hooke's law describes a certain light spring of unstretchedlength 36.0 cm. When one end isattached to the top of a door frame, and a 7.40-kg object is hung from the other end, the lengthof the spring is 49.00 cm. (a) Find its spring constant.(b) The load and the spring are taken down. Two people pull inopposite directions on the ends of the spring, each with a forceof 170 N. Find the length of thespring in this situation.
m (a) Find its spring constant.
(b) The load and the spring are taken down. Two people pull inopposite directions on the ends of the spring, each with a forceof 170 N. Find the length of thespring in this situation.
m
Explanation / Answer
Please remember to rate! -- -- Hooke's Law: F = -kx mg = -kx m =7.40kg g = 9.8 m/s^2 x = displacement = 49.00 - 36.0 = 13.0 cm = 0.13 m k = spring constant ---> negative gives direction, in thiscase down (7.40 kg)(9.8 m/s^2) = -k(0.13 m ) k = 72.52/0.13 k = 558 ----> not concerned about sign since weare calculating spring constant -- -- Part B: 170 N *2 = Force 340 N = 558(x) x = .61 m ----> not concerned about sign, since beingpulled from both ends -- -- Add this value to the unstretched length: Length = .36 m + .61 meter = .97m = 97.0 cm -- -- Hope this helps. Hooke's Law: F = -kx mg = -kx m =7.40kg g = 9.8 m/s^2 x = displacement = 49.00 - 36.0 = 13.0 cm = 0.13 m k = spring constant ---> negative gives direction, in thiscase down (7.40 kg)(9.8 m/s^2) = -k(0.13 m ) k = 72.52/0.13 k = 558 ----> not concerned about sign since weare calculating spring constant -- -- Part B: 170 N *2 = Force 340 N = 558(x) x = .61 m ----> not concerned about sign, since beingpulled from both ends -- -- Add this value to the unstretched length: Length = .36 m + .61 meter = .97m = 97.0 cm -- -- Hope this helps.Related Questions
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