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Two particles P and Q are attached to opposite ends of a light inextensible string which passes over a small smooth pulley at the top of a rough plane inclined at 30° to the horizontal. P has mass 0.2 kg and is held at rest on the plane. Q has mass 0.2 kg and hangs freely. The string is taut (see diagram). The coefficient of friction between P and the plane is 0.4. The particle P is released. Q strikes the floor and remains at rest. P continues to move up the plane for a further distance of 0.8 m before it comes to rest. P does not reach the pulley. Find the speed of the particles immediately before Q strikes the floor.

For this question I need to set 0.2a = 0.2gsin30 + 0.4x0.2gcos30. I don't really understand why we need to do this. Why couldn't I just use these simultaneous equations 0.2g - T = 0.2a and 0.2a = T- 0.2gsin30 -0.4x0.2gcos30? (The mark scheme says that this problem cannot be solved using these equations, but I don't understand why).

If you could provide any advice, I would really appreciate it.enter image description here

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    $\begingroup$ More appropriate for physics SE IMO $\endgroup$ Commented Oct 15, 2020 at 19:49
  • $\begingroup$ I don't know what you mean by SE IMO. Could you please explain? $\endgroup$
    – AOD
    Commented Oct 15, 2020 at 19:53

1 Answer 1

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Hint.

The friction work after the hanging ball strikes the floor is

$$ \mathcal{T}_f = \mu m_P g \cos\theta\Delta l $$

so we have

$$ \frac 12 m_P v_P^2 = \mathcal{T}_f + m_P g \Delta l \sin\theta $$

Here

$$ \cases{ \mu = 0.4\\ m_P = 0.2\\ \Delta l = 0.8\\ g = 9.81\\ \theta = 30^{\circ} } $$

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