A 1.0 kg object is suspended from a vertical spring whose spring constant is 130 N/m.
(a) Find the amount by which the spring is stretched from its unstrained length.
mg/k
To find the amount by which the spring is stretched, we need to use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is:
F = -kx
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement of the spring from its equilibrium position.
In this case, the weight of the object hanging on the spring is equal to the force exerted by the spring. The weight is given by:
W = mg
Where:
m is the mass of the object, and
g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth).
So, F = W = mg.
Substituting the values given in the problem, we have:
F = 1.0 kg * 9.8 m/s^2.
The force exerted by the spring is also given by Hooke's Law, so F = -kx. Therefore, we have:
-kx = 1.0 kg * 9.8 m/s^2.
Rearranging the equation, we can solve for x:
x = -F / k.
Substituting the values given in the problem, we have:
x = -(1.0 kg * 9.8 m/s^2) / 130 N/m.
Calculating this expression gives us the amount by which the spring is stretched from its unstrained length.