a 0.20 kg mass is attached to the end of a spring causes it to stretch 3cm. what is the spring constant?
65.3
Well, to find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. So, we have the equation: F = k * x, where F is the force, k is the spring constant, and x is the displacement.
Given that the mass is 0.20 kg and the displacement is 3 cm (which is 0.03 m), we can use Newton's second law, F = m * a, to find the force acting on the mass.
Now, since the mass is attached to the spring, the displacement is caused by the weight of the mass, which is given by W = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2). So, the force can be calculated as F = W = m * g.
Substituting this into Hooke's law equation, we get:
m * g = k * x
Now we can rearrange the equation to solve for the spring constant:
k = (m * g) / x
Plugging in the values, we get:
k = (0.20 kg * 9.8 m/s^2) / 0.03 m
So, the spring constant is approximately 65.33 N/m. But hey, don't forget to give the spring a break, it's been quite stretched!
To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.
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 mass (m = 0.20 kg) and the displacement (x = 3 cm = 0.03 m). We can calculate the force exerted by the spring using the formula:
F = m * g
Where:
m is the mass, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = (0.20 kg) * (9.8 m/s^2)
F = 1.96 N
Now we can rearrange Hooke's Law to solve for the spring constant:
k = -F / x
k = -(1.96 N) / (0.03 m)
k ≈ -65.33 N/m
Since the spring constant (k) cannot be negative, we take its absolute value:
k ≈ 65.33 N/m
Therefore, the spring constant is approximately 65.33 N/m.
To find the spring constant (k), which represents the stiffness of the spring, you 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 equation for Hooke's Law is:
F = -kx
Where:
F is the force applied to the spring (in newtons)
k is the spring constant (in newtons per meter)
x is the displacement or stretch of the spring (in meters)
In this case, you have a mass attached to the spring, so you need to consider the force due to gravity acting on the mass as well. The force due to gravity is given by:
Fg = mg
Where:
m is the mass (in kilograms)
g is the acceleration due to gravity (approximately 9.8 m/s²)
In equilibrium, when the spring is not stretched or compressed, the force due to gravity is balanced by the force exerted by the spring. Therefore:
Fg = Fs
Where:
Fs is the force exerted by the spring (in newtons)
Combining the above equations, you get:
mg = -kx
Rearranging the equation to solve for k, you have:
k = -(mg / x)
Now you can substitute the given values into this equation:
m = 0.20 kg
g = 9.8 m/s²
x = 0.03 m (since 3 cm equals 0.03 m)
k = -((0.20 kg) * (9.8 m/s²)) / 0.03 m
Simplifying this equation will give you the value of the spring constant, k.