a 3kg particle is in a region where the potential energy can be represented by t

Question

a 3kg particle is in a region where the potential energy can be represented by the function U(x)=(9/x^2)-4, where using x in meters will give you U in J. The particle is released from rest at x=-1.0 m

a) In which direction does it move? Why?

b) What is its velocity when it has moved 2.5 m?

c) describe the long term motion of the particle.

d) Consider now a different situation in which the particle is initially at x=3m and moving at a velocity of -4 m/s, what will its x-position be when it stops?

Explanation / Answer

F(x) = - dU/dx
F(x) = - d(9x^-2 - 4)/dx
F(x) = 18/x^3

(a)
Now, At x = -(1)
Force Acting on the Particle will be ,
F(-1) = -18 N

Therefore it will move in the -ve Direction.

(b)
Potential Energy at x = - 1.0 m
U(x) = 9-4 = 5J
Kinetic Energy at x = -1.0 m = 0

Potential Energy at x = - 3.5 m
U(x) = 9/3.5^2 - 4 = -3.26 J

Kinetic Energy at x = -3.5 m = ?

0 + 5 = - 3.26 + K.E
K.E = 8.26
1/2 * m*v^2 = 8.26
v^2 = (8.26 * 2)/ 3
v = 2.35 m/s

(c)
Force will Decrease & Eventualy Acceleration will decrease & particle will come to stop.

(d)
x = 3m
v = - 4m/s

Initial P.E = 9/3^2 - 4 = - 3J
Initial K.E = 1/2 * mv^2 = 1/2 * 3 * 4^2 = 24 J

When it Stops K.E = 0

Initial K.E + Final K.E = Initial K.E + Final P.E
24 - 3 = 0 + 9/x^2 - 4
25 = 9/x^2
x = 3/5
x = 0.6
X Position when it stops, x = 0.6

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