Potential Energy as a function of x

AI Thread Summary
The discussion focuses on calculating potential energy as a function of position x, given a force of 8e^{-2x} and an initial potential energy of 5 at x = 0. The integration of the force leads to the expression for work, W(x) = -4e^{-2x} + C, where C is a constant determined by the initial conditions. After solving for C, the correct potential energy function is derived as U(x) = 1 - 4e^{-2x}. The thread concludes with the realization that the integration limits were initially overlooked, leading to confusion in the calculations.
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[SOLVED] Potential Energy as a function of x

Homework Statement


Take U = 5 at x = 0 and calculate potential energy as a function of x, corresponding to the force:
8e^{-2x}


Homework Equations


W_{net} = U_i - U_f
W = \int_a^b F_x dx



The Attempt at a Solution



\int 8e^{-2x} dx = -4e^{-2x} = W(x)

W(x) = -4e^{-2x} = U_i(x) - U_f(x)
U_f(x) = U_i(x) + 4e^{-2x}
U_f(x) = 5 + 4e^{-2x}

correct answer: U(x) = 1 + 4e^{-2x}

Any ideas how i went wrong?
 
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You were working with definite integral, I guess. when you integrated 8exp(-2x), you should have put the limits (x = 0 and x = x).
 
SOLVED

W(x) = \int 8e^{-2x} = -4e^{-2x} + C ( I forgot the C earlier)
W(x) = U_i(x) - U_f(x)
-4e^{-2x} + C = 5 - U_f(x)
U_f(x) = 5 - 4e^{-2x} - C

SOLVE FOR C

U(0) = 5 = 5 - 4(1) - C
C = 4

SOLVE FOR U(x)

U_f(x) = 5 - 4e^{-2x} - 4

U_f(x) = 1 - 4e^{-2x}
 
Sourabh N said:
You were working with definite integral, I guess. when you integrated 8exp(-2x), you should have put the limits (x = 0 and x = x).

I am going to try this way as well when I get back from eating. Cheers!
 
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