You exert a 100-N pull on the end of a spring. When you increase the force by 20% to 120 N, the spring's length increases 4.0cm beyond its original stretched position.
What is the spring constant of the spring?
what about
What is its original displacement?
What is it's original displacement?
100=k x
100=(500 N/m) x
x= .2 M
To find the spring constant of the spring, we 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.
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 of the spring from its equilibrium position.
Given that you exert a 100 N force on the spring, we can substitute these values into the equation:
100 N = -kx₁
Next, we are given that when the force is increased to 120 N, the spring's length increases by 4.0 cm beyond its original stretched position. We can calculate the displacement of the spring using:
x₂ = x₁ + 4.0 cm
We can substitute this value and the new force into the equation:
120 N = -kx₂
Now, we have two equations:
100 N = -kx₁
120 N = -kx₂
To find the spring constant, we can rearrange the equations:
k = -100 N / x₁
k = -120 N / x₂
Now, we can substitute the values and solve for the spring constant:
k = -100 N / x₁ = -100 N / 0
k = -120 N / x₂ = -120 N / 0.04 m
Both calculations result in dividing by zero, which is not possible. This suggests that there might be an error in the given information, as it is not physically possible for the displacement to be zero or the spring constant to be infinite.
To find the spring constant accurately, we will need additional information regarding the force and displacement.
100 = k x
120 = k (x+.04) = k x + .04 k
-----------------------------subtract
-20 = kx - kx - .04 k
4 k = 2000
k = 500 N/m