Answer:
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
Step-by-step 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!!!