Question

The only force acting on a 2.0-kg body as it moves along the x-axis is given by:

Fx (x) = (12 - 2.0 x), where Fx is in N and x is in m.
The speed of the body at x = +2.0 m is 5.5 m / s.
What is the kinetic energy of this body when it is located at x = +6.0 m?

Answer 1

Step-by-step explanation:

To find the kinetic energy of the body at x = +6.0 m, we first need to determine the body's velocity at that position.

Given that the speed of the body at x = +2.0 m is 5.5 m/s, we can first find the acceleration at that point using Newton's second law:

F = m * a

Fx (x = 2.0 m) = (12 - 2.0 * 2.0) = 8 N

8 N = 2.0 kg * a

a = 4 m/s^2

Now, we can find the velocity at x = +6.0 m using kinematic equations. We assume the body starts from rest at x = 0:

v^2 = u^2 + 2 * a * s

where u is the initial velocity, a is the acceleration, and s is the displacement.

Since u = 0 and s = 6.0 m, we can calculate:

v^2 = 0 + 2 * 4 m/s^2 * 6.0 m

v^2 = 48 m^2/s^2

v = √(48) m/s

v ≈ 6.93 m/s

Now that we have the velocity at x = +6.0 m, we can calculate the kinetic energy using the formula:

KE = (1/2) * m * v^2

KE = (1/2) * 2.0 kg * (6.93 m/s)^2

KE ≈ 47.95 J

Therefore, the kinetic energy of the body at x = +6.0 m is approximately 47.95 Joules.