Relationship Between Conservative Forces and Potential Energy Problems and Solutions

 Problem#1

A single conservative force acting on a particle varies as F = (–Ax + Bx²)i, where A and B are constants and x is in meters. (a) Calculate the potential-energy function U(x) associated with this force, taking U = 0 at x = 0. (b) Find the change in potential energy and the change in kinetic energy as the particle moves from x = 2.00 m to x = 3.00 m.

Answer:
(a) the potential-energy function U(x) associated with this force, taking U = 0 at x = 0 is

U = –∫Fdx

U = –∫₀^x (–Ax + Bx²)dx

U = ½ Ax² – Bx³/3

(b) the change in potential energy and as the particle moves from x = 2.00 m to x = 3.00 m is

∆U = U_f – U_i

∆U = [½ A(3.00 m)² – B(3.00 m)³/3] – [½ A(2.00 m)² – B(2.00 m)³/3]

∆U = 2.50A – 19.0B/3

the change in kinetic energy as the particle moves from x = 2.00 m to x = 3.00 m is

∆K = –∆U = –2.50A + 19.0B/3

Problem#2
A single conservative force acts on a 5.00-kg particle. The equation F_x = (2x + 4) N describes the force, where x is in meters. As the particle moves along the x axis from x = 1.00 m to x = 5.00 m, calculate (a) the work done by this force, (b) the change in the potential energy of the system, and (c) the kinetic energy of the particle at x = 5.00 m if its speed is 3.00 m/s at x = 1.00 m.

Answer:
(a) the work done by this force is

W = ∫F_xdx

W = ∫₁^5.00m (2x + 4)dx

W = (x² + 4x)│₁^{5.00 m} = 40.0 J

(b) the change in the potential energy of the system is

∆U + ∆K = 0

∆U = –∆K = –W = –40.0 J

(c) the kinetic energy of the particle at x = 5.00 m if its speed is 3.00 m/s at x = 1.00 m.

∆K = K_f – K_i = K_f – ½ mv²

40.0J = K_f – ½ (5.00 kg)(3.00 m/s)²

K_f = 62.5 J

Problem#3
A potential-energy function for a two-dimensional force is of the form U = 3x³y – 7x. Find the force that acts at the point (x, y).

Answer:
the force that acts at the point (x, y) given by

F = F_xi + F_yj

With
F_x = –dU/dx = –d(3x³y – 7x)/dx = –9x²y + 7 and

F_y = –dU/dy = –d(3x³y – 7x)/dy = –3x³

Then
F = (–9x²y + 7)i – (3x³)j

Problem#4
The potential energy of a system of two particles separated by a distance r is given by U(r) = A/r, where A is a constant. Find the radial force F_r that each particle exerts on the other.

Answer:
the radial force F_r that each particle exerts on the other is

F_r = –dU/dr = –d(Ar^-1)/dr

F_r = A/r². The positive value indicates a force of repulsion.    

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