A particle moves along the axis under the influence of a variable force = 6.8x^2 + 4.4x where the force is measured in Newtons and the distance in meters. What is the potential energy associated with this force at = 4.0 m? Assume that = 0 J at = 0 m
Wouldn't you integrate fdx from zero to 4?
To find the potential energy associated with this force at a distance of 4.0 meters, we can integrate the force function over the given range and then evaluate it at 4.0 meters.
First, let's find the potential energy function U(x) by integrating the given force function F(x) with respect to x:
U(x) = ∫ F(x) dx
Given F(x) = 6.8x^2 + 4.4x, we can integrate term by term:
U(x) = ∫ (6.8x^2 + 4.4x) dx
To integrate 6.8x^2, use the power rule for integration:
∫ x^n dx = (1/(n + 1)) * x^(n + 1) + C
Integrating 6.8x^2:
∫ 6.8x^2 dx = (6.8/(2 + 1)) * x^(2 + 1) + C = 2.27x^3 + C1
To integrate 4.4x, use the power rule for integration:
∫ x^n dx = (1/(n + 1)) * x^(n + 1) + C
Integrating 4.4x:
∫ 4.4x dx = (4.4/(1 + 1)) * x^(1 + 1) + C = 2.2x^2 + C2
Now we can write the potential energy function U(x) as the sum of these integrals:
U(x) = 2.27x^3 + 2.2x^2 + C
Next, evaluate U(x) at x = 4.0 meters:
U(4.0) = 2.27(4.0)^3 + 2.2(4.0)^2 + C
Simplifying:
U(4.0) = 2.27(64) + 2.2(16) + C
U(4.0) = 145.28 + 35.2 + C
Finally, since the potential energy is defined relative to a reference point, we are given that U(0) = 0 J. Therefore, we can solve for the constant C by setting U(0) equal to zero:
U(0) = 2.27(0)^3 + 2.2(0)^2 + C
0 = 0 + 0 + C
C = 0
Now, we can find the potential energy at x = 4.0 meters:
U(4.0) = 145.28 + 35.2
U(4.0) = 180.48 J
Therefore, the potential energy associated with the force at a distance of 4.0 meters is 180.48 Joules.