Question

A 4.0-kg particle moving along the x-axis is acted upon by the force whose functional form appears below. The velocity of the particle at x=0 is v=6.0m/s.

Find the particle’s speed at x=(a)2.0m,
(b)4.0m,
(c)10.0m,
(d) Does the particle turn around at some point and head back toward the origin?
(e) Repeat part (d) if v=2.0m/satx=0.

Answer 1

Using work-energy principles, the speed of a 4.0-kg particle acted upon by a force F(x) = -cx³ can be determined at various positions and whether it turns around can also be established depending on its initial velocity.

Step-by-step explanation:

The question is about a 4.0-kg particle moving along the x-axis under the influence of a varying force described by the equation F(x) = -cx³ where c = 8.0 N/m³. To find the particle's speed at different positions along the x-axis, you need to apply work-energy principles.

The work done by the force as the particle moves from position x = 0 to any other position x will change its kinetic energy. Knowing the initial speed at x=0, and calculating the work done along the path, you can determine the speed at subsequent positions such as x = 2.0 m, x = 4.0 m, and x = 10.0 m.

By examining the direction of the force and the particle's kinetic energy, you can infer whether the particle turns around and heads back toward the origin. If the speed at x = 0 is changed to v = 2.0 m/s, the particle's subsequent motion can be re-evaluated to determine if and when it turns around.