Work is done only when there is a component of force in the direction of displacement. The applied force is vertical, while the block's displacement is horizontal. Since the force and displacement vectors are perpendicular to each other, the dot product of the two is zero, and no work is done by the vertical force.

The force exerted by the spring is $$F_s = -kx$$. The work done by the spring is the integral of this force over the displacement: $$W_s = \int_0^A (-kx) dx = -k\left[\frac{x^2}{2}\right]_0^A = -\frac{1}{2}kA^2$$. The work is negative because the spring force opposes the direction of displacement.

The normal force exerted by the ramp on the object is, by definition, perpendicular to the surface of the ramp. The object's displacement is parallel to the surface of the ramp. Since the normal force vector is always perpendicular to the displacement vector, the dot product is zero, and the work done by the normal force is zero.

The net work is the sum of the work done by the applied force, gravity, and friction. Work by applied force: $$W_F = (60 \text{ N})(10 \text{ m}) = 600 \text{ J}$$. Work by gravity: $$W_g = -mgh = -mgd\sin\theta = -(5)(10)(10)\sin(30^\circ) = -250 \text{ J}$$. Normal force: $$N = mg\cos\theta = (5)(10)\cos(30^\circ) \approx 43.5 \text{ N}$$. Work by friction: $$W_f = -f_k d = -\mu_k N d = -(0.20)(43.5)(10) = -87 \text{ J}$$. Net work: $$W_{net} = 600 - 250 - 87 = 263 \text{ J}$$.

The work-energy theorem states that the net work done on a particle is equal to the change in its kinetic energy ($$W_{net} = \Delta K$$). Since kinetic energy is $$K = \frac{1}{2}mv^2$$, it is always non-negative. If the net work is negative, the change in kinetic energy is negative ($$\Delta K < 0$$), which means the final kinetic energy is less than the initial kinetic energy. This implies that the particle's speed must decrease.

For Path 1, the line is $$y=x$$, so $$dy=dx$$. $$\vec{F} = k(x\hat{i} - x\hat{j})$$ and $$d\vec{r} = dx(\hat{i} + \hat{j})$$. $$W_1 = \int_0^L k(x-x)dx = 0$$. For Path 2, the first leg is from (0,0) to (L,0), where $$y=0, dy=0$$. $$W_{2a} = \int_0^L k(-x\hat{j}) \cdot(dx\hat{i}) = 0$$. The second leg is from (L,0) to (L,L), where $$x=L, dx=0$$. $$W_{2b} = \int_0^L k(y\hat{i} - L\hat{j}) \cdot(dy\hat{j}) = \int_0^L -kL dy = -kL^2$$. Total work $$W_2 = W_{2a} + W_{2b} = -kL^2$$. Thus, $$W_1=0$$ and $$W_2=-kL^2$$.

The instantaneous rate at which work is done is power, given by $$P = \vec{F} \cdot \vec{v}$$. First, find the velocity vector by taking the derivative of the position vector: $$\vec{v}(t) = \frac{d\vec{r}}{dt} = (2t \hat{i} + 2 \hat{j})$$ m/s. At $$t=1$$ s, the velocity is $$\vec{v}(1) = (2 \hat{i} + 2 \hat{j})$$ m/s. Now, calculate the power: $$P = (5 \hat{i} - 3 \hat{j}) \cdot(2 \hat{i} + 2 \hat{j}) = (5)(2) + (-3)(2) = 10 - 6 = 4$$ W.

The path can be parameterized by $$y=2x$$, so $$dy=2dx$$. The displacement vector is $$d\vec{r} = dx \hat{i} + dy \hat{j} = dx \hat{i} + 2dx \hat{j}$$. Along this path, the force is $$\vec{F} = (2(2x)\hat{i} + 3x\hat{j}) = (4x\hat{i} + 3x\hat{j})$$. The work is $$W = \int \vec{F} \cdot d\vec{r} = \int_0^2 (4x dx + (3x)(2dx)) = \int_0^2 10x dx = \left[5x^2\right]_0^2 = 5(4) = 20 \text{ J}$$.

This question tests understanding of work in physics, specifically the calculation of work done by a force applied at an angle. Work is defined as the product of force and displacement in the direction of the force, calculated using W = Fd cos(θ). In this scenario, you are given a force of 18 N applied at an angle of 25° above the horizontal, and a horizontal displacement of 4.0 m. Choice C is correct because it accurately applies the work formula: W = (18 N)(4.0 m)cos(25°) = 72 × 0.906 = 65.2 J ≈ 65 J. Choice A incorrectly calculates 18 × 4 = 72 J without considering the angle. When teaching this concept, emphasize drawing free body diagrams to visualize the force components. Have students practice identifying which angle to use in the cosine function - it's always the angle between the force vector and displacement vector.

This question tests understanding of work when force is perpendicular to displacement. Work is defined as W = Fd cos(θ), where θ is the angle between force and displacement vectors. In this scenario, a 15 N force is applied at 90° to the 10 m displacement direction. Choice B is correct because cos(90°) = 0, giving W = (15 N)(10 m)(0) = 0 J. Choices A and D incorrectly calculate as if the force were parallel or antiparallel to displacement. This is a fundamental concept: perpendicular forces do no work because they don't contribute to motion in the displacement direction. Use examples like circular motion where centripetal force is always perpendicular to velocity, doing no work. Have students practice identifying perpendicular force-displacement pairs in various contexts.