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!!!

i would say

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft