The potential energy of a particle moving in a certain one-dimensional region of

Question

The potential energy of a particle moving in a certain one-dimensional region of space is U(x) = (1.00 J/m4)x4 (4.00 J/m2)x2 + (1.00 J/m)x.
You might want to plot the function U(x) using a graphing calculator or an online utility such as desmos.com. An object of mass 2.00 kg is released from rest at x = 2.00 m.

(a) At what position does the object have maximum speed?

(b) What is the maximum speed of the object?

(c) What is the velocity of the object when it reaches the unstable equilibrium point?

(d) What is the lowest value of x that the object gets to?

Explanation / Answer

a) Maximum speed is at a point where potential energy is minimum. From graph point is x=-1.473

b) for maximum speed we use concept mechanical energy is conserved
U2 + K2 = U-1.473 + K-1.473
2 + 0 = -5.444 + K-1.473
K-1.473 = 7.444

c) Unstable equilibrium is local maxima at x=0.126, with corresponding PE = 0.063
Again applying mechanical energy conservation betweet starting point and x=0.126, we get
2+0 = 0.063 + K0.126
K0.126 = 1.937

d) Lowest value of x, object gets to is point where potential energy = initial PE, that makes KE at that pont = 0

From graph that point is x= -2.25

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