Hooke's law describes a certain light spring of unstretched length 38.0 cm. When one end is attached to the top of a door frame and a 6.00-kg object is hung from the other end, the length of the spring is 42.0 cm. a. Find its spring constant.
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) 1470 N/m

(b) 48.2 m

Explanation:

Applying,

(a) F = ke Equation 1

Where F = force applied to the spring, k = spring constant, e = extension

make k the subject of the equation

k = F/e Equation 2

But,

F = mg Equation 3

Where m = mass, g = acceleration due to gravity

Substitute equation 3 into equation 2

k = mg/e Equation 4

From the question,

Given: m = 6 kg, e = 42-38 = 4 cm = 0.04 m

Constant: g = 9.8 m/s²

Substitute these values into equation 4

k = (6×9.8)/0.04

k = 1470 N/m

(b) Consider the end of the spring to the left which exert a force to the right

Then,

e = F/k Equation 5

Given: F = 150 N, k = 1470 N/m

Substitute these values into equation 5

e = 150/1470

e = 0.102 m

Hence the length of the spring is

L = 0.38+0.102 = 0.482 cm = 48.2 m

we need answer choices

f=ma newton

f=100x2.5=250n