Variable force with integration (force as function of position)

A particle moves under a variable force that depends on position (often inversely proportional to distance squared), requiring integration of F·dx to find work done or using work-energy principle.

2 questions · Challenging +1.2

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CAIE M2 2018 November Q4
8 marks Challenging +1.2
\includegraphics{figure_4} A particle \(P\) of mass \(0.5\text{ kg}\) is projected along a smooth horizontal surface towards a fixed point \(A\). Initially \(P\) is at a point \(O\) on the surface, and after projection, \(P\) has a displacement from \(O\) of \(x\text{ m}\) and velocity \(v\text{ m s}^{-1}\). The particle \(P\) is connected to \(A\) by a light elastic string of natural length \(0.8\text{ m}\) and modulus of elasticity \(16\text{ N}\). The distance \(OA\) is \(1.6\text{ m}\) (see diagram). The motion of \(P\) is resisted by a force of magnitude \(24x^2\text{ N}\).
  1. Show that \(v\frac{\text{d}v}{\text{d}x} = 32 - 40x - 48x^2\) while \(P\) is in motion and the string is stretched. [3]
  2. The maximum value of \(v\) is \(4.5\). Find the initial value of \(v\). [5]
CAIE M2 2018 November Q4
8 marks Challenging +1.2
\includegraphics{figure_4} A particle \(P\) of mass \(0.5\text{ kg}\) is projected along a smooth horizontal surface towards a fixed point \(A\). Initially \(P\) is at a point \(O\) on the surface, and after projection, \(P\) has a displacement from \(O\) of \(x\text{ m}\) and velocity \(v\text{ m s}^{-1}\). The particle \(P\) is connected to \(A\) by a light elastic string of natural length \(0.8\text{ m}\) and modulus of elasticity \(16\text{ N}\). The distance \(OA\) is \(1.6\text{ m}\) (see diagram). The motion of \(P\) is resisted by a force of magnitude \(24x^2\text{ N}\).
  1. Show that \(v\frac{\text{d}v}{\text{d}x} = 32 - 40x - 48x^2\) while \(P\) is in motion and the string is stretched. [3] The maximum value of \(v\) is \(4.5\).
  2. Find the initial value of \(v\). [5]