This question tests understanding of elastic potential energy with a stiff spring and small compression. Elastic potential energy is calculated using U = ½kx², where the quadratic dependence on displacement means even small compressions in stiff springs can store significant energy. In this scenario, we have k = 500 N/m and x = 0.0800 m, giving U = ½(500)(0.0800)² = ½(500)(0.0064) = 1.60 J. Choice A is correct because it properly applies the formula with the squared displacement term. Choice B (3.20 J) is incorrect because it omits the ½ factor, a common error when students confuse the spring force formula with the energy formula. To help students: Emphasize the quadratic relationship between displacement and energy - doubling the compression quadruples the stored energy. Practice with various spring constants to build intuition about how stiffness affects energy storage.

This question tests understanding of elastic potential energy in spring systems, specifically calculating the energy stored when a spring is compressed. Elastic potential energy is given by the formula U = ½kx², where k is the spring constant and x is the displacement from equilibrium. In this scenario, we have a spring with k = 200 N/m compressed by x = 0.150 m, requiring us to substitute these values into the formula. Choice A is correct because U = ½(200)(0.150)² = ½(200)(0.0225) = 2.25 J. Choice B (4.50 J) is incorrect because it results from forgetting the ½ factor in the formula, a common mistake when students confuse the spring force formula (F = kx) with the energy formula. To help students: Emphasize that potential energy formulas often include a ½ factor due to integration, and practice dimensional analysis to verify that the units work out to joules. Create visual aids showing the parabolic relationship between compression distance and stored energy.

This question tests understanding of gravitational potential energy, specifically how to calculate it using mass, height, and gravitational acceleration. Gravitational potential energy is the energy stored by an object due to its position in Earth's gravitational field, calculated using U = mgh. In this scenario, we have an object with mass m = 2.00 kg at height h = 3.00 m, with g = 9.80 m/s², requiring direct substitution into the formula. Choice A is correct because U = (2.00)(9.80)(3.00) = 58.8 J, properly multiplying all three values. Choice C (29.4 J) is incorrect because it results from using half the correct value, possibly confusing gravitational potential energy with kinetic energy's ½mv² formula. To help students: Stress that gravitational potential energy has no ½ factor unlike elastic potential energy, and practice identifying which energy formula to use based on the physical situation. Use real-world examples like lifting objects to reinforce the concept that more mass or greater height requires more work.

This question tests understanding of potential energy changes during downward motion on an incline. When an object descends, its gravitational potential energy decreases regardless of the path taken - only the vertical height change matters. In this scenario, a 1.20 kg block descends a vertical height of 2.50 m, so ΔU = mgΔh = (1.20)(9.80)(-2.50) = -29.4 J. Choice A is correct because it properly includes the negative sign indicating a decrease in potential energy. Choice B (29.4 J) is incorrect because it ignores the direction of motion, treating the energy change as positive when the object is actually losing potential energy. To help students: Emphasize that the incline angle is irrelevant for potential energy calculations - only vertical displacement matters. Use energy conservation to show that this lost potential energy becomes kinetic energy, making the negative sign physically meaningful.

This question tests understanding of potential energy changes in gravitational systems, specifically when an object descends and loses height. When an object moves downward in a gravitational field, its potential energy decreases, resulting in a negative change: ΔU = mgΔh, where Δh is negative for downward motion. In this scenario, a 4.00 kg block descends 1.50 m, so Δh = -1.50 m, giving ΔU = (4.00)(9.80)(-1.50) = -58.8 J. Choice A is correct because it properly accounts for the negative change in height and the resulting negative change in potential energy. Choice B (58.8 J) is incorrect because it ignores the direction of motion, treating the descent as if it were an ascent, a common sign error students make. To help students: Emphasize that potential energy changes depend on the direction of motion - decreasing height means decreasing potential energy. Use energy conservation examples to show that lost potential energy becomes kinetic energy, reinforcing the negative sign's physical meaning.