What is the spring constant of a spring that stretches 2.00 cm when a mass of 0.600 kg is suspended from it?
Answer
i know the answer is 295 n/m but im nto sure how they got to it,
Spring Constant =
Weight(N) divided by stretch (meters)
= 0.60*9.8/0.02 = 294 N/m
i feel like slapping myself across the face... i completely forgot about gravity...
Why did the spring go to therapy? Because it had tension issues! But don't worry, I'll help you with your question.
To find the spring constant (k) in this case, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be expressed as F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the displacement (x) is given as 2.00 cm, which we need to convert to meters by dividing by 100, giving us 0.020 m.
The force (F) can be calculated using the equation F = mg, where m is the mass (0.600 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = (0.600 kg)(9.8 m/s^2) = 5.88 N
Now, substituting the values into Hooke's Law, we get:
5.88 N = -k(0.020 m)
To solve for the spring constant (k), we divide both sides by -0.020 m:
k = 5.88 N / -0.020 m
And if you evaluate that, you'll find that the spring constant is indeed approximately 295 N/m.
I hope that springed some clarity into the situation!
To find the spring constant of a spring, you can use Hooke's Law, which 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 applied by the spring, k is the spring constant, and x is the displacement of the spring.
In this case, the mass of 0.600 kg is suspended from the spring and causes it to stretch by 2.00 cm (which can be converted to meters by dividing by 100).
First, let's calculate the force exerted by the spring using the formula F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = (0.600 kg)(9.8 m/s^2) = 5.88 N
Now, we can use Hooke's Law to find the spring constant:
-5.88 N = -k(0.02 m)
Dividing both sides by -0.02 m, we get:
k = 5.88 N / 0.02 m = 294 N/m
So, the spring constant of the spring is 294 N/m.
To find the spring constant of a spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement 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 from the equilibrium position.
In this case, we are given the displacement, x = 2.00 cm = 0.02 m, and the mass, m = 0.600 kg. We want to find the spring constant, k.
First, we need to calculate the force exerted by the spring using the equation:
F = mg
Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
F = (0.600 kg)(9.8 m/s^2)
F = 5.88 N
Now, we can rearrange Hooke's Law to solve for the spring constant:
F = -kx
k = -F/x
Plugging in the values we have:
k = -(5.88 N) / (0.02 m)
k = -294 N/m
Note that the negative sign indicates that the force exerted by the spring is in the opposite direction of its displacement.
Therefore, the spring constant of the spring is approximately 294 N/m.