To streatch a certain nonlinear spring by an amount 2x requires a force F given by F=40x-6x^2 where F is in newtons and x is in meters. What is the change in potential energy when the spring is stretched 2 meters from its equilibrium position?
a)16 J
b)28 J
c)56 J
d)64 J
e)80 J
P.E. = mgh but how would I relate this to this current problem? Or would the answer just be c) 56 J ( I got this by plugging in 2 for x in the equation)
PE= INT F dx
integrated F dx from zero to 2m.
To find the change in potential energy when the spring is stretched 2 meters, we'll need to integrate the force function with respect to distance.
The potential energy, PE, stored in the spring can be calculated by integrating the force, F, with respect to distance, x:
PE = ∫ F dx
In this case, the force function is given as F = 40x - 6x^2.
Integrating F with respect to x:
PE = ∫ (40x - 6x^2) dx
To evaluate this integral, we'll need to split it into two parts:
PE = ∫ 40x dx - ∫ 6x^2 dx
Let's start by evaluating the first integral:
∫ 40x dx = 20x^2 + C1
Now, let's evaluate the second integral:
∫ 6x^2 dx = 2x^3 + C2
Combining the two integrals:
PE = 20x^2 + C1 - 2x^3 + C2
Since we're interested in the change in potential energy when the spring is stretched 2 meters from its equilibrium position, we'll evaluate the integral from x = 0 to x = 2:
PE = [20(2)^2 + C1 - 2(2)^3 + C2] - [20(0)^2 + C1 - 2(0)^3 + C2]
PE = 80 + C1 - 16 + C2 - (0 + C1 - 0 + C2)
PE = 80 - 16 + (C1 - C1) + (C2 - C2)
PE = 80 - 16 + 0 + 0
PE = 64 J
Therefore, the change in potential energy when the spring is stretched 2 meters from its equilibrium position is 64 J (option d).