Question

A force F(x) = (3.4/x) N acts on a particle as it moves along the positive x-axis. (a) How much work (in J) does the force do on the particle as it moves from x = 2.4 m to x = 6.0 m? 3.115 J (b) Picking a convenient reference point of the potential energy to be zero at x = 1 m, find the potential energy for this force. (Use the following as necessary: x. Assume U(x) is in joules and x is in meters. Do not include units in your answer.) U(x) = -9.069

          A force F(x) = (3.4/x) N acts on a particle as it moves along the positive x-axis.
(a) How much work (in J) does the force do on the particle as it moves from x = 2.4 m to x = 6.0 m?
3.115 J
(b) Picking a convenient reference point of the potential energy to be zero at x = 1 m, find the potential energy for this force. (Use the following as necessary: x. Assume U(x) is in joules and x is in meters. Do not include units in your answer.)
U(x) = -9.069
        
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A force F(x) = (3.4/x) N acts on a particle as it moves along the positive x-axis.
(a) How much work (in J) does the force do on the particle as it moves from x = 2.4 m to x = 6.0 m?
3.115 J
(b) Picking a convenient reference point of the potential energy to be zero at x = 1 m, find the potential energy for this force. (Use the following as necessary: x. Assume U(x) is in joules and x is in meters. Do not include units in your answer.)
U(x) = -9.069

Added by Lorenzo K.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Transcript

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00:01 Suppose we have a force acting on a particle which is given as, we'll write as 3 .4 newton meters over x and we want to know what is the work done if it goes from an initial position of 2 .4 meters to 6 meters.
00:19 So for part a the work done is just going to be the integral from xi to xf of f of x dx and so this is going to be 3 .4 joules...
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