An object with mass m moves in one dimension subject to the potential energy U (

Question

An object with mass m moves in one dimension subject to the potential energy U ( x) = ax^2 + bx^4 . Here both a and b are greater than 0. The mechanical energy of the system is E = 2a^2/b . There are no numerical values given in this problem so the answers should be expressed in terms of a and b.

(a) Derive an expression for the positions x = ±x0 away from the origin where the force on the object vanishes.

(b) Write an expression for the maximum kinetic energy of the object Kmax .

(c) Write an expression for the maximum distance xmax from the origin that can be reached given E. (d) Determine the force Fmax and its direction when x = xmax .

(e) Provide graphs that show the kinetic energy and the force on the object versus x and provide a qualitative verbal description of the motion of the object.

Explanation / Answer

a)

U ( x) = ax2 + bx4

F(x) = - dU(x) /dx = - (-2ax + 4bx3)

when F(x) = 0

2 a x - 4bx3 = 0

x = sqrt(a/2b)

b)

Kinetic energy is maximum when potential energy is 0

so KInetic energy = mechanical energy - potential energy

Kinetic energy = 2a2/b

c)

E = ax2 + bx4

ax2 + bx4 = 2a2/b

bx4 ax2 = 2a2/b

b2x4 abx2 = 2a2

x = sqrt (2a/b)

d)

F(x) = (2ax - 4bx3)

dF(x)/dx = 2a - 12bx2

for max. force , dF(x)/dx = 0

2a - 12bx2 = 0

x = sqrt (a/6b)

Fmax = (2ax - 4bx3) = 2 a ( sqrt (a/6b)) - 4 b (a/6b)3/2

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