As long as you don’t stretch it too much, the force exerted by the spring obeys

Question

As long as you don’t stretch it too much, the force exerted by the spring obeys Hooke’s Law. The further you stretch the spring, the harder it pulls back on you. The stiffness of a spring is represented by the spring constant (see Section 10.1 in the text). In this problem consider the spring to be of negligible mass.
Suppose you stretch a spring 0.0769 m from its equilibrium position. The spring has a spring constant of 1.26×103 N/m. What is the magnitude of the force you need to exert on the spring?
96.9 N

Suppose you place this same spring on a hook and hang a 0.396 kg object on the end. How far does the spring stretch from equilibrium? (think about the force that the object exerts on the spring).

cannot figure out number 2

Explanation / Answer

Stretched length of spring x = 0.0769 m

Spring constant K = 1.26 x 103 N/m

(a) Force F = kx

F = (1.26 x 103) (0.0769)

F = 96.9 N

(b) If we hang a mass vertically then its weight acts on the spring

F = mg

that will stretch the spring

so mg = kx

0.396 * 9.8 = ( 1.26 x 103) x

x = 3.08 x 10-3 m

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