Hooke\'s law describes a certain light spring of unstretched length37.0 cm. When

Question

Hooke's law describes a certain light spring of unstretched length37.0 cm. When one end is attached tothe top of a door frame, and a 8.00-kgobject is hung from the other end, the length of the spring is41.50 cm. (a) Find its spring constant.
1 kN/m

(b) The load and the spring are taken down. Two people pull inopposite directions on the ends of the spring, each with a force of150 N. Find the length of the springin this situation.
2 m (a) Find its spring constant.
1 kN/m

(b) The load and the spring are taken down. Two people pull inopposite directions on the ends of the spring, each with a force of150 N. Find the length of the springin this situation.
2 m

Explanation / Answer

Hooke's Law: F = -kx mg = -kx m = 8.00 kg g = 9.8 m/s^2 x = displacement = 41.50 - 37.0 = 4.5 cm = 0.045 m k = spring constant ---> negative gives direction, in thiscase down (8.00 kg)(9.8 m/s^2) = -k(0.045 m ) k = 78.4/0.045 k = 1742 ----> not concerned about sign since we arecalculating spring constant Part B: 150 N *2 = Force 300 N = 1742(x) x = .1722m ----> not concerned about sign, sincebeing pulled from both ends Add this value to the unstretched length: Length = .37 m + .1722 meter = .542 m = 54.2 cm Hope this helps.

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