Hooke’s Law

Section 1.10
66 (a-b

Use Hooke’s Law, which states that the distance a spring stretches (or compresses) from its natural, or equilibrium, length varies directly as the applied force on the spring.


66. A force of 265 newtons stretches a spring 0.15 meter.

(a) What force stretches the spring 0.1 meter?
(b) How far does a force of 90 newtons stretch the spring?

Let L = length
Let k = constant of proportionality
Let f = applied force

L = fk

I gotta first find k.

0.15 = 265k

0.15/265 = k

0.0005660377 = k

Part (a)

I gotta find f.

0.1 = f(0.0005660377)

0.1 ÷ 0.0005660377 = f

176.6666666667 = f

I will round to two decimal places.

176.67 newtons = f

Part (b)

I need to find L.

L = 90(0.0005660377)

L = 0.050943393

Rounding to two decimal places I get

L = 0.05 meters

You say?

 

correct

 

I got another word problem right. It's a miracle.