Source: Principles of Physics
Problem
We move a particle along an x axis, first outward from $x = 1.0$ m to $x = 4.0$ m and then back to $x = 1.0$ m, while an external force acts on it. That force is directed along the x axis, and its x component can have different values for the outward trip and for the return trip. Here are the values (in newtons) for four situations, where x is in meters:
Outward / Inward (a) +3.0 / -3.0 (b) +5.0 / +5.0 (c) +2.0x / -2.0x (d) +3.0x^2 / +3.0x^2
[QUESTIONS] [Q(a-d)] Find the net work done on the particle by the external force for the round trip for each of the four situations. [Q(e)] For which, if any, is the external force conservative? [/QUESTIONS]
(a) $W_{net} = 18.0$ J (b) $W_{net} = 0$ J (c) $W_{net} = 30.0$ J (d) $W_{net} = 0$ J (e) The forces in (b) and (d) are conservative.
The net work done for a round trip is the sum of the work done on the outward trip ($W_{out}$) and the inward trip ($W_{in}$). Work is calculated by the integral $W = \int F(x) dx$. The outward path is from $x_i = 1.0$ m to $x_f = 4.0$ m, and the inward path is from $x_f = 4.0$ m to $x_i = 1.0$ m. A force is conservative if the net work done over a closed path is zero.
(a) $F_{out} = +3.0$ N, $F_{in} = -3.0$ N
$$W_{net} = \int_{1.0}^{4.0} 3.0 \,dx + \int_{4.0}^{1.0} (-3.0) \,dx$$ $$W_{net} = 3.0[x]_{1.0}^{4.0} - 3.0[x]_{4.0}^{1.0} = 3.0(4.0 - 1.0) - 3.0(1.0 - 4.0) = 9.0 \text{ J} + 9.0 \text{ J} = 18.0 \text{ J}$$(b) $F_{out} = +5.0$ N, $F_{in} = +5.0$ N
$$W_{net} = \int_{1.0}^{4.0} 5.0 \,dx + \int_{4.0}^{1.0} 5.0 \,dx$$ $$W_{net} = 5.0[x]_{1.0}^{4.0} + 5.0[x]_{4.0}^{1.0} = 5.0(4.0 - 1.0) + 5.0(1.0 - 4.0) = 15.0 \text{ J} - 15.0 \text{ J} = 0 \text{ J}$$(c) $F_{out} = +2.0x$ N, $F_{in} = -2.0x$ N
$$W_{net} = \int_{1.0}^{4.0} 2.0x \,dx + \int_{4.0}^{1.0} (-2.0x) \,dx$$ $$W_{net} = [x^2]_{1.0}^{4.0} - [x^2]_{4.0}^{1.0} = (4.0^2 - 1.0^2) - (1.0^2 - 4.0^2) = 15.0 \text{ J} + 15.0 \text{ J} = 30.0 \text{ J}$$(d) $F_{out} = +3.0x^2$ N, $F_{in} = +3.0x^2$ N
$$W_{net} = \int_{1.0}^{4.0} 3.0x^2 \,dx + \int_{4.0}^{1.0} 3.0x^2 \,dx$$ $$W_{net} = [x^3]_{1.0}^{4.0} + [x^3]_{4.0}^{1.0} = (4.0^3 - 1.0^3) + (1.0^3 - 4.0^3) = 63.0 \text{ J} - 63.0 \text{ J} = 0 \text{ J}$$(e) The forces in situations (b) and (d) are conservative because the net work done over the closed loop is zero.