Final answer:
To find the particle's speed at B (xB = -2.0 m), we applied the work-energy theorem, equating the work done by the force F(x) = -cx^3 during the particle's travel from A to B with the change in kinetic energy. By calculating the integral for work and rearranging the equation, we determine that the final speed vB is approximately 7.1 m/s, closest to option c) 8 m/s.
Step-by-step explanation:
The question involves using the work-energy principle to find the speed of a 4.0 kg particle constrained to move along the x-axis subject to force F(x) = -cx^3. We can use the work-energy theorem to equate the work done by this force over the distance from point A to point B to the change in kinetic energy of the particle. The work done by the force F(x) from point A to B is the negative of the integral of F(x) from xA to xB:
W = ∫xAxB F(x) dx = ∫1.0-2.0 (-8.0 x^3) dx = -∫1.0-2.0 (-8.0 x^3) dx = -8.0 [(-2.0)^4/4 - (1.0)^4/4] = -8.0 (4 - 1/4) = -30 J (negative because force is opposite to displacement)
Delta KE = KEB - KEA = (1/2 m vB^2) - (1/2 m vA^2)
Using the work-energy theorem (W = Delta KE), we get -30 J = (1/2 * 4.0 kg * vB^2) - (1/2 * 4.0 kg * 6.0 m/s^2)
Solving for vB gives us vB = sqrt((30 J * 2 + 4.0 kg * 6.0 m/s^2)/(4.0 kg))
Substitute the numbers and calculate: vB = sqrt((60 J + 144 J) / 4.0 kg) = sqrt(204 J / 4.0 kg) = sqrt(51) m/s
This gives us a final speed vB = approximately 7.1 m/s
Therefore, the particle's speed at B, where xB = -2.0 m, is approximately 7.1 m/s, which is closest to option c) 8 m/s, though none of the choices are an exact match.