A single conservative force F(x) acts on a 4.7 kg particle that moves along an x

Question

A single conservative force F(x) acts on a 4.7 kg particle that moves along an x axis. The potential energy U(x) associated with F(x) is given by
U(x) = -2.1xe-x/4
where x is in meters. At x = 2.0 m the particle has a kinetic energy of 6.2 J. (a) What is the mechanical energy of the system? (b) What is the maximum kinetic energy of the particle and (c) the value of x at which it occurs?

I got 3.65J for part A and 6.66J for part B which are both right.
I got 5m for part C, but it is wrong.
Please help! I am down to my last attempt.

Explanation / Answer

Mass of the particle m = 4.7 Kg

Potential energy U = -2.1x exp(-x/4)

(a)

At X = 2 m,

Kinetic energy K = 6.2 J

Potential energy U = -2.1*2 exp(-1/2) = -2.54 J

So total mechanical energy E = K + U = 6.2 - 2.54 = 3.65 J

(b)

Maximum Kinetic energy Kmax = E - U(at x = 4)

                                             = 3.65 - (-3.09) = 6.65 J

(c)

Maximum Kinetic energy occurs at x = 4 m  

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