MathsFurther Mechanics 2 › Newton's laws with a variable force

Newton's laws with a variable force

When the force depends on where the particle is, F = ma becomes a differential equation. Which form of the acceleration to use is decided entirely by what the force depends on.

Year FMEDEXCEL 9FM0 FM2

Builds on Work, energy and power and Solving differential equations.

IN THIS TOPIC

  • Choose between dv/dt and v dv/dx according to what the force depends on.
  • Solve the resulting differential equation and apply the initial conditions.
  • Handle inverse square forces, including work done against gravity.

WHAT YOU PROBABLY THINK

For a variable force you can use the SUVAT equations with the average value of the force.

Choosing the right acceleration

Acceleration can be written as dv/dt or as v dv/dx, and the two are equal by the chain rule. Which one to use is not a matter of taste: match it to the variable the force depends on, so that the resulting equation separates.

F(x) = mvdvdx,     F(t) = mdvdt

The SUVAT equations assume constant acceleration and are simply not available here, whatever average is used: the opening claim would give the wrong answer for every non-uniform force. Work-energy still applies, though, with the work found by integration, and it is often the quickest route when speeds are wanted at positions.

A force that falls off with distance: the work it does is the area under the graph, and it runs out at x = 5work = 42 J5 m20 NF = 20 − 4x2 kg from rest: v = 6.48 m/s at x = 3
FIG. 1A force falling off with distance, with the work done to a point given by the area under the graph.

WORKED EXAMPLE

A force that dies away

A particle of mass 2 kg starts from rest at the origin under a force F = 20 − 4x newtons. Find its speed at x = 3.

The force depends on x, so use 2v dv/dx = 20 − 4x.

Separating and integrating: v² = ∫(20 − 4x) dx = 20x − 2x² + c, and v = 0 at x = 0 gives c = 0.

At x = 3: v² = 60 − 18 = 42, so v = 6.48 m/s.

The same 42 J is the area under the force-distance graph, which is the work-energy principle saying the same thing.

Inverse square forces

Gravitation gives F = GMm/x² directed towards the centre. Rather than quoting G and M, use the fact that the force equals mg at the surface, where x = R, so GM = gR² and the force becomes mgR²/x². That removes both constants and keeps the arithmetic in terms of quantities the question supplies.

Because the force depends on position, use v dv/dx. Integrating gives the speed as a function of distance, and the work done moving from R to a distance d is the integral of the force, which comes out as mgR²(1/R − 1/d). Letting d grow without bound gives mgR, the energy needed to escape entirely.

An inverse square force: at twice the radius it is a quarter, and the work out to there is half mgRR2Rmgmg/4F = GMm/x²work out to 2R = mgR/2
FIG. 2The inverse square force falling to a quarter at twice the radius, with the work out to that point marked.

YOUR TURN

Climbing away from a planet

A body is projected vertically from the surface of a planet of radius R where the surface gravity is g. Find the work done against gravity in reaching a height R above the surface, per unit mass.

Show the working

The force per unit mass at distance x is gR²/x².

Work = ∫gR²/x² dx from R to 2R = gR²[−1/x] from R to 2R.

That is gR²(1/R − 1/2R) = gR/2.

Half the escape energy takes you only one radius up, which is why escape is so much harder than it first looks.

THE EXAM BIT

  • Decide between dv/dt and v dv/dx by looking at what the force depends on, and say why.
  • Never reach for SUVAT: state that the acceleration is not constant.
  • Apply the initial conditions immediately after integrating, before rearranging.
  • For gravitation, replace GM by gR² so that the constants disappear.

CHECK YOURSELF

A particle of mass 1 kg moves under a force F = 6x newtons, starting from rest at x = 1. Find its speed at x = 3.

Show a hint

The force depends on x, so use v dv/dx.

Show the answer

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d

v

/

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x

=

6

x

,

s

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v

²

=

6

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t

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v

=

0

a

t

x

=

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,

c

=

6

.

A

t

x

=

3

:

v

²

=

5

4

6

=

4

8

,

s

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v

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6

.

9

3

m

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.

Use v dv/dx when the force depends on position and dv/dt when it depends on time; SUVAT is unavailable either way.

For gravitation, write GM as gR² so the inverse square force is mgR²/x², and integrate it for the work done.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Choose between dv/dt and v dv/dx according to what the force depends on.
  • Solve the resulting differential equation and apply the initial conditions.
  • Handle inverse square forces, including work done against gravity.

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