The only force acting on a 2.20-kg body as it moves along a positive x axis has an x component Fx = -4.40x N, with x in meters. The velocity at x = 7.60 m is 8.59 m/s.

Question

The only force acting on a 2.20-kg body as it moves along a positive x-axis has an x component F_x = -4.40x N, with x in meters. The velocity at x = 7.60 m is 8.59 m/s.

(a) What is the velocity of the body at x = 3.27 m?

(b) At what positive value of x will the body have a velocity of v = 4.73 m/s?

Answer 1

Let

m = 2.20 kg be the mass of the body,

k = -4.40 N/m be the given stiffness,

F(x) = -kx be the force acting at position x,

v(x) be the velocity at position x.

We are given

v(7.6) = 8.59 m/s

You need to recognize the situation of

"F = -kx" as the situation of a spring with a spring constant k,

and recall the formulas for energy.

When the body is at a position x

the body has kinetic energy mv²(x)/2,

and the spring has potential elastic energy kx²/2.

So the total energy at position x is

E(x) = mv²(x)/2 + kx²/2. (1)

Given that there is no other force (such as friction), the energy is conserved.

That is it, is the same for every position x.

(a)

By conservation of energy

E(3.27) = E(7.6)

Substitute from (1)

mv²(3.27)/2 + k3.27²/2 = mv²(7.6)/2 + k7.6²/2

Express

v(3.27) = ± √(v²(7.6) + (k/m)(7.6² - 3.27²)) = ± √(8.59² + (4.4/2.2)(7.6² - 3.27²)) ≈ 13

There are two solution: At x = 3.27 m the velocity will be 13 m/s or -13 m/s.

During the oscillation when the body is moving in the positive x direction the velocity will be 13 m/s.

On the return trip it will be -13 m/s.

(b)

We are to find x satisfying

v(x) = 4.73

At that position x, by conservation of energy

E(x) = E(7.6)

Substitute from (1)

mv²(x)/2 + kx²/2 = mv²(7.6)/2 + k7.6²/2

Substitute the given velocities

m4.73²/2 + kx²/2 = m8.59²/2 + k7.6²/2

Express

x = ± √((m/k)(8.59² - 4.73²) + 7.6²) = ± √((2.2/4.4)(8.59² - 4.73²) + 7.6²) ≈ ± 9.1

Again there are two solutions:

The body will have velocity 4.73 m/s at position x = 9.1 m and also at x = -9.1 m.