Hooke's law describes a certain light spring of unstretched length 31.0 cm. when one end is attached to the top of a door frame, and a 7.60-kg object is hung from the other end, the length of the spring is 40.00 cm.
(a) find its spring constant.
1. kn/m

(b) the load and the spring are taken down. two people pull in opposite directions on the ends of the spring, each with a force of 150 n. find the length of the spring in this situation.

a) k= 827.5N/m

b) L = 0.67m

Explanation:

a)

we know that:

F = kx

where F is the force execute in the spring, k the spring constant and x the change in the length of the spring, so:

mg = kx

where m is the mass of the object and g is the gravity. Replacing values, we get:

(7.6kg*9.8) = K(0.4m-0.31m)

Solving for k:

k = 827.5N/m

b) First, we plus the forces of the two people as:

F = 150N + 150N

F = 300N

Finally, using the following equation:

F = Kx

300N = 827.5N/m(x)

Solving for x:

x = 0.36m

It means that the new length of the spring is:

L = 0.36m+0.31m

L = 0.67m

v=ir = 3x15 =45 v

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