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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?
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?