The only force acting on a 2.00 kg body as it moves along a positive x axis has an x component Fx = −kx, where k = 6.00 N/m. The velocity at x = 3.00 m is 8.00 m/s. (a) What is the velocity of the body at x = 4.00 m? (b) At what positive value of x will the body have a velocity of 5.00 m/s?

a) v at x = 4.0 m is 6.56 m/s

b) The body will have a velocity of 5.00 m/s at x = 4.69 m

Explanation:

Using the work-energy theorem, the work done in moving an object between two points is equal to the change in kinetic energy of the body between those two points.

W = ΔK.E

Note that workdone by a position-variable force is given as the definite integral between the two points of F.dx

W = ∫ F.dx (integral evaluated from point 1 to point 2)

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Given

F = -kx = -6x

Mass of body = 2.00 kg

Velocity at x = 3.0 m is 8.0 m/s

(a) The velocity of the body at x = 4.00 m?

Workdone by the force moving from point x=3 to x=4

W = ∫⁴₃ Fdx = ∫⁴₃ (-6x) dx = [-3x²]⁴₃

= [-3(4²)] - [-3(3²)]

= -48 - (-27)

= -21 J

W = ΔK.E = -21

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = ?

Initial kinetic energy at x=3

= (½)(2)(8²) = 64 J

ΔK.E = (final kinetic energy) - (initial kinetic energy)

-21 = (final kinetic energy) - 64

final kinetic energy = -21 + 64 = 43 J

Final kinetic energy = (½)(2)(v²) = 43

v² = 43

v = 6.557 m/s

(b) At what positive value of x will the body have a velocity of 5.00 m/s?

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Initial kinetic energy at x=3

= (½)(2)(8²) = 64 J

Final kinetic energy = (½)(2)(5²) = 25 J

ΔK.E = 25 - 64 = - 39 J

W = ΔK.E = - 39 J

W = ∫ˣ₃ Fdx = ∫ˣ₃ (-6x) dx = [-3x²]ˣ₃

= [-3(x²)] - [-3(3²)]

W = -3x² + 27 = - 39

-3x² = -39-27 = -66

x² = 22

x = 4.69 m

Hope this Helps!!!

high-sugar diet and high cholesterol