A new type of spring is found where the spring force F is given by F=-kx + bx^2 where k=400 N/ m and b=100 N/m^2 and x is the expansion (or compression) of the spring from its equilibrium position. The spring is attached to a point mass of 1 kg sitting on a horizontal frictionless table.

A. Find he work done by the spring acting on the attached mass when the mass is pulled by a person from x=0 to x=-0.5m.
B. If the mass is released from rest by the person at the -0.5 m point, what is the speed of the mass at x=0?

To find the work done by the spring acting on the attached mass, we need to calculate the integral of the spring force with respect to displacement (x).

A. The work done can be found using the following equation:

W = ∫F dx

Given that the spring force F is given by F = -kx + bx^2, we can substitute this into the equation:

W = ∫(-kx + bx^2) dx

To find the limits of integration, we are told that the mass is pulled from x=0 to x=-0.5m. Therefore, the lower limit will be 0 and the upper limit will be -0.5.

W = ∫(-kx + bx^2) dx
= ∫(-kx) dx + ∫(bx^2) dx
= (-k/2) ∫x^2 dx + (b/3) ∫x^3 dx

Now, let's calculate each integral:

∫x^2 dx = (x^3)/3
∫x^3 dx = (x^4)/4

Plugging these back into the equation:

W = (-k/2) * [(x^3)/3] + (b/3) * [(x^4)/4]

Evaluate the expression at the limits of integration (0 and -0.5):

W = (-k/2) * [((-0.5)^3)/3] + (b/3) * [((-0.5)^4)/4]

Substituting the values for k and b:

W = (-400/2) * [((-0.5)^3)/3] + (100/3) * [((-0.5)^4)/4]

Now, calculate this expression to find the work done.

B. To find the speed of the mass at x=0 when it is released from rest at x=-0.5m, we can use the principle of conservation of mechanical energy.

The initial mechanical energy (Ei) will be the sum of the gravitational potential energy (PEi) and the potential energy due to the spring (PEs):

Ei = PEi + PEs

At x=0, the gravitational potential energy is zero since the mass is at the same height as the reference point. Therefore, PEi = 0.

The potential energy due to the spring at x=0 is given by:

PEs = (1/2) kx^2

To find the final mechanical energy (Ef), we only need to consider the kinetic energy (KEf) at x=0 since there is no potential energy present:

Ef = KEf

According to the principle of conservation of mechanical energy, Ei = Ef. Therefore:

PEs = KEf

Substitute the expression for the potential energy due to the spring and solve for the kinetic energy at x=0:

(1/2) kx^2 = KEf

(1/2) * 400 * (0.5)^2 = KEf

Now, we can calculate the kinetic energy using this equation. The speed of the mass at x=0 can be found by taking the square root of the calculated kinetic energy.

A. To find the work done by the spring, we need to integrate the force with respect to displacement.

The work done by the spring is given by the equation:

W = ∫ F dx

First, let's plug in the values for k, b, and the displacement limits:

W = ∫ (-kx + bx^2) dx

W = ∫ (-400x + 100x^2) dx

Now, we can integrate term by term:

W = [-200x^2 + (100/3)x^3] evaluated from x = 0 to x = -0.5

W = [-200(-0.5)^2 + (100/3)(-0.5)^3] - [-200(0)^2 + (100/3)(0)^3]

W = [-200(0.25) + (100/3)(-0.125)] - [0]

W = -50 + (-4.17)

W ≈ -54.17 Joules

Therefore, the work done by the spring acting on the attached mass when it is pulled from x = 0 to x = -0.5m is approximately -54.17 Joules.

B. To find the speed of the mass at x = 0 when it is released from rest at x = -0.5m, we can use the principle of conservation of mechanical energy.

The total mechanical energy, E, is the sum of the kinetic energy and the potential energy:

E = KE + PE

At x = -0.5m, the potential energy is:

PE = 0.5kx^2 - 0.33bx^3

PE = 0.5(400)(-0.5)^2 - 0.33(100)(-0.5)^3

PE = 100 - 4.17

PE ≈ 95.83 Joules

At x = 0, the potential energy is:

PE = 0.5kx^2 - 0.33bx^3

PE = 0.5(400)(0)^2 - 0.33(100)(0)^3

PE = 0 Joules

Since the table is frictionless, the total mechanical energy is conserved:

E = KE + PE

E = 0.5mv^2 + 0

95.83 Joules = 0.5(1)(v^2)

v^2 = 191.66

v ≈ 13.86 m/s

Therefore, the speed of the mass at x = 0 is approximately 13.86 m/s.