GRE Physics Companion: Conservative Forces and Potential Energy

GRE Physics Companion: Conservative Forces and Potential Energy

The fastest way to approach conservative-force questions is to recognize which relation is being tested.

The four central relations are

(1)

(2)

(3)

and, near Earth’s surface,

(4)

1 High-value GRE facts

1. Positive conservative work means potential energy decreases.
2. Negative conservative work means potential energy increases.
3. A conservative force does zero net work around a closed path.
4. Potential energy is defined only up to an additive constant.
5. In one dimension,

   (5)

6. Force points toward decreasing potential energy.
7. A local minimum of U is stable equilibrium.
8. A local maximum of U is unstable equilibrium.
9. For an ideal spring,

   (6)

10. Near Earth’s surface,

    (7)

11. With U(∞) = 0, Newtonian gravitational potential energy is

    (8)

Part I: Original GRE-style problems

Problem 1: sign of potential-energy change

A conservative force does +40 J of work on a particle. The particle’s potential energy changes by

1. −40 J
2. −20 J
3. 0
4. +20 J
5. +40 J

Problem 2: closed path

A particle moves around a closed loop under a conservative force. The net work done by that force is

1. always negative
2. zero
3. equal to the maximum potential energy
4. equal to the path length times the average force
5. impossible to determine

Problem 3: force from potential

The potential energy of a particle is

(9)

in SI units. The force is

1. +6x
2. +3x
3. 0
4. −3x
5. −6x

Problem 4: gravitational potential-energy change

A 2.0 kg mass rises by 5.0 m near Earth’s surface. Taking g = 9.8 m∕s², the change in gravitational potential energy is

1. −98 J
2. −49 J
3. 0
4. +49 J
5. +98 J

Problem 5: spring energy

An ideal spring has spring constant k = 200 N∕m and is compressed by 0.10 m. Its elastic potential energy relative to equilibrium is

1. 0.5 J
2. 1.0 J
3. 2.0 J
4. 10 J
5. 20 J

Problem 6: stable equilibrium

A one-dimensional potential-energy curve has an equilibrium at x = x₀. Which condition identifies stable equilibrium?

1. U′(x₀) > 0
2. U′(x₀) < 0
3. U′(x₀) = 0 and U′′(x₀) > 0
4. U′(x₀) = 0 and U′′(x₀) < 0
5. U(x₀) = 0

Problem 7: path independence

A mass moves from height h₁ to lower height h₂ under uniform gravity. Path A is vertical and Path B is a long curved track. Neglecting all forces except gravity, which statement is correct?

1. Gravity does more work along Path A.
2. Gravity does more work along Path B.
3. Gravity does the same work along both paths.
4. Gravity does zero work along both paths.
5. The result depends on the travel time.

Problem 8: additive constant

If

(10)

where C is constant, the force obtained from U′ is

1. larger than the original force by C
2. smaller than the original force by C
3. identical to the original force
4. zero
5. undefined

Problem 9: gravitational potential

With the convention U(∞) = 0, the Newtonian gravitational potential energy of masses M and m separated by r is

1. +GMm∕r
2. −GMm∕r
3. +GMm∕r²
4. −GMm∕r²
5. 0

Problem 10: slope of a potential curve

At some point on a one-dimensional potential-energy graph,

(11)

The force at that point points

1. in the positive x direction
2. in the negative x direction
3. perpendicular to the x axis
4. nowhere because the force is zero
5. in a direction that cannot be inferred

Problem 11: work from spring potential

A spring with k = 100 N∕m moves from extension x_i = 0.30 m to x_f = 0.10 m. The work done by the spring is

1. −4.0 J
2. −2.0 J
3. 0
4. +2.0 J
5. +4.0 J

Problem 12: double-well equilibrium

For

(12)

which statement is correct?

1. x = 0 is the only equilibrium and it is stable.
2. x = 0 is the only equilibrium and it is unstable.
3. x = 0 is stable and two additional equilibria are unstable.
4. x = 0 is unstable and two additional equilibria are stable.
5. There are no equilibrium points.

Part II: Complete worked solutions

Solution 1

For a conservative force,

(13)

Thus

(14)

Answer: (A).

Solution 2

A conservative force satisfies

(15)

Therefore the net work around any closed path is zero.

Answer: (B).

Solution 3

Use

(16)

Since

(17)

we have

(18)

Thus

(19)

Answer: (E).

Solution 4

Near Earth’s surface,

(20)

Therefore

ΔUg	= (2.0)(9.8)(5.0)	(21)
	= 98 J.	(22)

Answer: (E).

Solution 5

The elastic potential energy is

(23)

Thus

Us	= (200)(0.10)2	(24)
	= 1.0 J.	(25)

Answer: (B).

Solution 6

Stable equilibrium occurs at a local minimum of potential energy:

(26)

Answer: (C).

Solution 7

Uniform gravity is conservative. Its work depends only on the initial and final heights:

(27)

Therefore both paths give the same gravitational work.

Answer: (C).

Solution 8

The force from U′ is

Fx′	= −	(28)
	= −(U + C)	(29)
	= −.	(30)

The constant disappears upon differentiation.

Answer: (C).

Solution 9

With

(31)

Newtonian gravity has

(32)

Answer: (B).

Solution 10

Since

(33)

a negative slope means

(34)

The force points in the positive x direction.

Answer: (A).

Solution 11

The spring work is

(35)

Therefore

Ws	= (100)(0.30)2 −(100)(0.10)2	(36)
	= 4.5 − 0.5	(37)
	= 4.0 J.	(38)

Answer: (E).

Solution 12

Differentiate:

(39)

Factor:

(40)

The equilibrium points are

(41)

and

(42)

The second derivative is

(43)

At the origin,

(44)

so x = 0 is unstable.

At the other two equilibria,

(45)

so they are stable.

Answer: (D).

2 GRE checklist

Before calculating, identify which representation is shortest.

1. If conservative work is given, use ΔU = −W_c.
2. If U(x) is given, use F_x = −dU∕dx.
3. If the force is given, integrate −F_x dx to obtain U.
4. If gravity near Earth’s surface appears, use ΔU_g = mgΔy.
5. If a spring appears, use U_s = kx².
6. If an equilibrium curve is shown, use the slope and curvature of U(x).
7. If two different paths connect the same endpoints under a conservative force, the work is the same.
8. Do not assign physical meaning to the absolute zero of potential energy unless a reference convention has been stated.

References

References