Question

The only force acting on a 2.0kg body as it moves along a positive x-axis has an x-component Fx=-6x with x in meters.The velocity at x =3.0m is 8.0m/s.The velocity of the body at x=4.0m is?

Answer 1

equation for the force is given as

$F_{x}$ = - 6x

$x_{i}$ = initial position = 3 m

$x_{f}$ = final position = 4 m

work done by a force is given as

W = $\int_{x_{i}}^{x_{f}}$ $F_{x}$ dx

W = $\int_{3}^{4}$ (- 6x) dx

W = - 3 ((4²) - (3)²)

W = - 21 J

m = mass of the body = 2 kg

v₁ = initial velocity = 8 m/s

v₂ = final velocity = ?

Using work-change in kinetic energy theorem

Work done = change in kinetic energy

Work done = final Kinetic energy - initial kinetic energy

W = (0.5) m (v²₂ - v²₁)

- 21 = (0.5) (2) (v²₂ - 8²)

v₂ = 6.6 m/s

Answer 2

Explanation:

hope it helps u.....