Spring-like force acts on a body

Question

The only force acting on a 3.50-kg body as it moves along a positive x axis has an x component F_x = -6.40x N, with x in meters. The velocity at x = 4.82 m is 7.35 m/s.

a) What is the velocity of the body at x = 1.59 m?

b) At what positive value of x will the body have a velocity of v = 4.41 m/s?

Answer 1

While Arturo was entering his answer, I was working out the numbers as well:

The initial information gives us the total Mechanical Energy of the system, composed of elastic potential energy and kinetic energy:  ME = 1/2 k x² + 1/2 m v². Because spring force = -kx, the spring constant k is 6.40 N/m. So initial ME = 1/2 (6.4) (4.82)² + 1/2 (3.5) (7.35)² = 168.88 J.

a) Because there is no friction or other force, ME is conserved. Now ME = 1/2 (6.4) (1.59)² + 1/2 (3.5) v² = the same 168.88 J. Then 1.75 v² = 168.88 - 8.09, v² = 160.79/1.75 = 91.88, and v = 9.59 m/s.

b) Again ME is conserved, 1/2 (6.4) x² + 1/2 (3.5) (4.41)² = 168.88. So 3.2 x² = 168.88 - 34.03, x² = 134.85/3.2 = 42.14, and x = 6.49 m.

Answer 2

(a)

Use energy conservation.

kx₁² / 2 + mv₁² / 2 = kx₂² / 2 + mv₂² /2 ⇒

v₂ = {(2/m) [k(x₁² - x₂²) / 2 + mv₁² / 2]}^1/2

Plug in the numbers and get v₂:

m = 3.50 kg

k = 6.4 N/m

x₁ = 4.82 m

v₁ = 7.35 m/s

x₂ = 1.59 m

(b)

Use the total energy E from part (a). It will be the same in part (b).

E = kx₁² / 2 + mv₁² / 2

Evaluate E using the same numbers as in part (a). Then

E = kx₃² / 2 + mv₃² / 2 ⇒

x₃ = [(2/k) (E - mv₃² / 2)]^1/2

Plug in k, E, and m from part (a), plug in and v₃ = 4.41 m/s, and get x₃ in meters.