Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by
[tex] U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2} [/tex]
a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00
[tex] u(x) = \frac{-2}{x} + \frac{4}{x^2}[/tex]
At x = 1.00 m
[tex] u(1) = \frac{-2}{1} + \frac{4}{1^2}[/tex]
= 4J
Kinetic energy = (1/2)mv²
[tex] = \frac{1}{2} * 4(3)^2 [/tex]
= 18J
Total energy will be =
4J + 18J = 22J
At x = 5
[tex] u(5) = \frac{-2}{5} + \frac{4}{5^2}[/tex]
[tex] = \frac{4-10}{25} = \frac{-6}{25} J[/tex]
= -0.24J
Kinetic energy =
[tex] \frac{1}{2} * 4Vf^2 [/tex]
= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2
[tex] Vf = \sqrt{frac{22.4}{2}[/tex]
= 3.33 m/s
b) magnitude of force when x = 5.0m
[tex] u(x) = \frac{-2}{x} + \frac{4}{x^2}[/tex]
[tex] \frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}[/tex]
[tex] = \frac{2}{x^2} - \frac{8}{x^3}[/tex]
At x = 5.0 m
[tex] \frac{2}{5^2} - \frac{8}{5^3} [/tex]
[tex] F = \frac{2}{25} - \frac{8}{125}[/tex]
= 0.016N
