A particle of mass 2 kg, starting from the rest at the origin, moves under the action of a force, which at time t is given by F = 8 cos 2ti + 8 sin 2tj. In the example on page 383, it was shown that at time t, the velocity is v = 2 sin 2ti + (2 - 2 cos 2t)j the acceleration is

Question

A particle of mass 2 kg, starting from the rest at the origin, moves under the action of a force, which at time t is given by

F = 8 cos 2ti + 8 sin 2tj.

In the example on page 383, it was shown that at time t, the velocity is

v = 2 sin 2ti + (2 - 2 cos 2t)j

the acceleration is

a = 4 cos 2ti + 4 sin 2tj

and that the kinetic energy is

E = 8(1 - cos 2t).

i) Find the first time t (t > 0) that the velocity is perpendicular to the acceleration.
ii) Find the rate of working at time t.
iii) Verify that the power is equal to the rate of change of kinetic energy.

Accepted Answer

i) When the acceleration is perpendicular to the velocity, [latex]a \cdot v = 0.[/latex]

[latex]a \cdot v [/latex]= (a cos 2t) (2 sin 2t) + (4 sin 2t)(2 - 2 cos 2t)

= 8 sin 2t cos 2t + 8 sin 2t - 8 sin 2t co 2t

= 8 sin 2t

The first time for which [latex]a \cdot v = 0[/latex] is thus when sin 2t = 0, that is t = π/2.

ii) Power = [latex]F \cdot v [/latex] = [8 cos 2ti + 8 sin 2tj] [latex]\cdot[/latex] [2 sin 2ti + (2 - 2 cos 2t)j]

= 16 sin 2t cos 2t + 8 sin 2t (2 - 2 cos 2t)

= 16 sin 2t

iii) Given the kinetic energy E = 8(1 - cos 2t), the rate of change of kinetic energy is

[latex] \begin{matrix} \frac{dE}{dt} = 16 sin 2t & \boxed{ ext{This is a general result. Work done by all forces = increase in kinetic energy Rate of working = power = rate of increase in kinetic energy}} \end{matrix}[/latex]

which is the power.

Question 17.6: A particle of mass 2 kg, starting from the rest at the origi...

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