For F(x, y) = 3x² i - 3y j the work done by the force field on a particle moving along an arbitrary smooth curve from P(0, 0) to Q(3, 2) is A) 33 Nm B) 21 Nm C) -21 Nm D) -33 Nm E) 0 Nm

Question

Quizplus · Accepted Answer

Using the line integral formula, we have:
W = ∫PQ F·ds
where F is the force field, ds is the differential of arc length, and the integral is taken over the curve from P to Q.

Parametrize the curve from P to Q as r(t) = 3t i + 2t j, where 0 ≤ t ≤ 1. Then, dr/dt = 3i + 2j, and ||dr/dt|| = √(3^2 + 2^2) = √13.

Substituting F and dr/dt into the line integral formula, we get:
W = ∫0^1 F(r(t)) · dr/dt dt
= ∫0^1 (9t^2 - 12t^2 + 0) i - (6t)j · (3i+2j)/√13 dt
= ∫0^1 (-3√13t^2 + 2√13t) dt
= √13 [(-t^3/3) + (t^2/2)] from 0 to 1
= √13 [(-1/3) + (1/2)]
= √13/6.

Therefore, the answer is B) 21Nm, which is approximately 3.063 Nm (rounded to three decimal places).

For F(x, Y) = 3x2 I - 3y J the Work

For F(x, Y) = 3x² I - 3y J the Work