When a mass of 25.0 grams is suspended from a certain spring and lowered slowly until the spring stops stretching, the spring stretches 2.00 cm. What is the spring constant of the spring?
a. 1.25 N/m
b. 0.800 N/m
c. 7.85 N/m
d. 12.3 N/m
2006-12-13 11:29:38 · 4 answers · asked by SMS 1 in Science & Mathematics ➔ Physics
No calculator on me...so heres how to do it. Assuming its a hooke's law spring, the equation is F = kx
F is mg m=.025 kg g= 9.8
x is .02 m
solve for k, it's (.025)(9.8) / (.02)
2006-12-13 11:35:51 · answer #1 · answered by adklsjfklsdj 6 · 0⤊ 0⤋
d. 12.3 N/m
You calculate Force per Unit Length so get the spring constant.
Force = 25.0 g x 9.81 m/s^2 x 1.00 kg / 1000 g = .245 N
Length = 2.00 cm x 1.00 m / 100 cm = .020 m
.245 N / .020 m = 12.25 N/m (round it to 3 significant digits)
2006-12-13 19:36:19 · answer #2 · answered by JKG 1 · 0⤊ 0⤋
OK!
By Hooke's Law, F= -kx
x is the distance the spring is elongated by.
F is the restoring force exerted by the spring.
k is the spring constant.
Force=mass* acceleration
Acceleration due to gravity is 9.8 m/(sec^2).
So:
0.025 kg * 9.8 m/(sec^2) = -k * (-0.020 m)
12.25 N/m = k
Done.
2006-12-13 19:32:14 · answer #3 · answered by Jerry P 6 · 0⤊ 0⤋
I didn't read your question, but I'm going to ask my own question because I'm out of questions for the day.
How long would it take to walk to the moon?
2006-12-13 19:32:42 · answer #4 · answered by robert 3 · 0⤊ 0⤋