A particle of mass 1 kg moves along a line with a velocity v ms^-1 under the influence of a resistive force of magnitude kv² N, where k is a constant. Initially, the velocity of the particle is 10 ms^-1 and the force continues to act until the particle has slowed down to 5 ms^-1. i) Use kinetic

Question

A particle of mass 1 kg moves along a line with a velocity v [latex]ms^{-1}[/latex] under the influence of a resistive force of magnitude kv² N, where k is a constant.
Initially, the velocity of the particle is 10 [latex]ms^{-1}[/latex] and the force continues to act until the particle has slowed down to 5 [latex]ms^{-1}[/latex].
i) Use kinetic energy considerations to write down the work done by the resistive force.
ii) Solve the equation of motion of the particle and express the displacement x in terms
of v.
Hence show that the particle travels a distance [latex]s = \frac{1}{k}[/latex] ln 2 m while the force acts.
iii) From part ii) express v and hence F in terms of x.
Hence confirm by integration the result obtained in part i).

Accepted Answer

i) Work done = final K.E. - initial K.E.

[latex]= \frac{1}{2} × 5² - \frac{1}{2} × 10² J[/latex]

= - 37.5 J.

This is negative because the force is in the opposite direction to the displacement.

ii) Applying F = ma with m = 1 and a = [latex]v\frac{dv}{dx},[/latex] where x is displacement,

[latex]-kv² = v\frac{dv}{dx}.[/latex]

Separating the variables gives

[latex]-k\int{dx} = \int{\frac{dv}{v}}[/latex]

⇒ -kx = lnv + c.

When x = 0, v = 10, hence c = -ln 10. So

-kx = lnv - ln 10

= - ln[latex]\frac{10}{v}[/latex]

⇒ x = [latex]\frac{1}{k} ln \frac{10}{v}[/latex] ①

When v = 5 (final velocity)

x = s = [latex]\frac{1}{k}[/latex] ln2.

iii) Making v the subject in ① gives v = 10 [latex]e^{-kx}[/latex]. Therefore

F = -kv²

= -100k[latex]e^{-2kx}[/latex]

Work done is [latex]\int{F dx}:[/latex]

[latex]\int{F dx} = \int^{s}_{0}{-100k e^{-2kx} dx}[/latex]

= [latex]50\int^{s}_{0}{-2k e^{-2kx} dx}[/latex]

= [latex]50\left[e^{-2kx}\right]^{s}_{0}[/latex]

Now [latex]s = \frac{1}{k} ln 2 [/latex] so

[latex]e^{-2ks} = e^{-2 ln 2}[/latex]

= [latex]e^{ln \frac{1}{4}}[/latex]

⇒ [latex]e^{-2ks} = \frac{1}{4}.[/latex]

Thus the work done = 50 ([latex]\frac{1}{4} - 1[/latex])

= - 37.5.

This is -37.5 J, as in part i).

Question 18.5: A particle of mass 1 kg moves along a line with a velocity v...

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