GRE Physics Companion: Work by Variable Forces

GRE Physics Companion: Work by Variable Forces

Variable-force work problems are usually testing one of four ideas:

1. recognizing work as a signed area under an F_x versus x graph,
2. evaluating a simple integral,
3. using Hooke’s law correctly,
4. combining variable-force work with W_net = ΔK.

The central formulas are

(1)

and, in one dimension,

(2)

1 High-value GRE facts

1. Area under an F_x versus x graph has units of joules.
2. Area under an F_x versus t graph has units of newton-seconds, not joules.
3. Regions below the x axis on an F_x versus x graph contribute negative work.
4. A linear force-position segment is usually fastest to handle with triangle or trapezoid area.
5. For a spring, the force is F_x = −kx.
6. Work done by a spring from x_i to x_f is

   (3)

7. If a slowly applied external force stretches a spring from 0 to x, the external work is +kx².
8. Reversing the same path reverses the sign of the work.
9. In more than one dimension, the path can matter.
10. The fastest route to speed is often

    (4)

Part I: Original GRE-style problems

Problem 1: linear force graph

A force increases linearly from 0 at x = 0 to 12 N at x = 4.0 m. The work done from x = 0 to x = 4.0 m is

1. 12 J
2. 18 J
3. 24 J
4. 36 J
5. 48 J

Problem 2: positive and negative graph areas

From x = 0 to x = 6 m, the area between an F_x versus x curve and the axis is +30 J above the axis and 12 J below the axis. The net work is

1. −42 J
2. −18 J
3. 18 J
4. 30 J
5. 42 J

Problem 3: power-law force

A one-dimensional force is

(5)

The work done from x = 0 to x = L is

1. aL
2. aL²∕2
3. aL³∕3
4. aL³
5. 2aL³

Problem 4: spring work

An ideal spring has k = 200 N∕m. What work does the spring do as it moves from x = 0.20 m to equilibrium?

1. −8.0 J
2. −4.0 J
3. 0
4. 4.0 J
5. 8.0 J

Problem 5: stretching a spring farther

A spring with constant k is stretched slowly from x to 2x. The work done by the external agent during this part of the stretch is

1. kx²
2. kx²
3. kx²
4. 2kx²
5. 4kx²

Problem 6: work-energy with variable force

A 2.0 kg particle starts from rest at x = 0 under the net force

(6)

in SI units. Its speed at x = 2.0 m is closest to

1. 2.0 m∕s
2. 3.5 m∕s
3. 4.9 m∕s
4. 6.0 m∕s
5. 12 m∕s

Problem 7: average force

A force increases linearly with position from 4 N to 10 N over a displacement of 3.0 m. The work is

1. 7 J
2. 14 J
3. 18 J
4. 21 J
5. 30 J

Problem 8: identifying the correct graph

Which graph has an area that directly equals mechanical work in one-dimensional motion?

1. acceleration versus time
2. velocity versus time
3. force versus time
4. force versus position
5. momentum versus time

Problem 9: path dependence

A force field is

(7)

A particle moves from (0, 0) to (L,L).

Path A goes first along x at y = 0, then vertically at x = L. Path B goes first vertically to y = L, then horizontally to x = L.

Which statement is correct?

1. Both paths give zero work.
2. Both paths give aL².
3. Path A gives 0 and Path B gives aL².
4. Path A gives aL² and Path B gives 0.
5. The work is undefined because the endpoints are different.

Problem 10: reversing a path

A force does +15 J of work as a particle moves along a specified path from A to B. If the same geometric path is traversed from B back to A while the same force field applies, the work is

1. −30 J
2. −15 J
3. 0
4. +15 J
5. +30 J

Part II: Complete worked solutions

Solution 1

The force-position graph is a triangle:

(8)

Answer: (C).

Solution 2

Signed areas add:

(9)

Answer: (C).

Solution 3

Integrate:

W	= ∫ 0Lax2 dx	(10)
	= a0L	(11)
	= .	(12)

Answer: (C).

Solution 4

The spring work is

(13)

Here

(14)

Thus

Ws	= (200)(0.20)2	(15)
	= 4.0 J.	(16)

The spring does positive work while returning toward equilibrium.

Answer: (D).

Solution 5

For a slow stretch, the external work from x₁ to x₂ is

(17)

With

(18)

we obtain

Wext	= k(2x)2 −kx2	(19)
	= 2kx2 −kx2	(20)
	= kx2.	(21)

Answer: (C).

Solution 6

The net work is

Wnet	= ∫ 026xdx	(22)
	= 02	(23)
	= 12 J.	(24)

Since the particle starts from rest,

(25)

Therefore

(26)

so

(27)

Answer: (B).

Solution 7

For a linearly varying force, the average force is the arithmetic mean:

(28)

Therefore

(29)

Answer: (D).

Solution 8

In one dimension,

(30)

Thus the signed area under a force-versus-position graph is work.

Answer: (D).

Solution 9

For Path A, the horizontal segment lies at y = 0, so

(31)

The vertical segment is perpendicular to the force. Therefore

(32)

For Path B, the first vertical segment again gives zero work. The top horizontal segment lies at y = L, so

(33)

Hence

(34)

Answer: (C).

Solution 10

Reversing the same path changes

(35)

Therefore

(36)

Answer: (B).

2 GRE checklist

Before integrating, ask whether geometry is faster.

1. Is the force-position graph made of rectangles, triangles, or trapezoids?
2. Have I included the sign of areas below the axis?
3. If the force is a simple function, can I integrate it directly?
4. If a spring appears, did I use F = −kx and distinguish work by the spring from work on the spring?
5. If speed is requested, can I immediately use W_net = ΔK?
6. If the problem is multidimensional, has the path been specified?
7. Am I integrating with respect to position rather than time when calculating work?

References

References