This a quite simple question, unfortunately I can't get why my solution and the correct solution don't match.

A particle P of mass $0.2$ kg rests on a rough plane inclined at $30$ degrees to the horizontal. The coefficient of friction between the particle and the plane is $0.3$. A force of magnitude $0.25$N acts on P up the plane, parallel to a line of greatest slope of the plane. Starting from rest, P slides down the plane. After moving a distance of $3\,$m, P passes through the point A. Find speed at A.

The solution is

$0.3 × 0.2\mathrm g \cos 30 × 3\ [= 1.5588\, \mathrm J] $
(WD against F = friction × distance)

$WD = 0.25 × 3\ [= 0.75 \, \mathrm J] $ (WD against 0.25 force)

$0.2g × 3 \sin 30\ [= 3 \, \mathrm J] $(PE loss = mgh)

$\left[{1 \over 2} (0.2) v^2 = 3 – 1.5588 – 0.75\right]$ (Work/Energy equation)

Speed = $2.63\, \mathrm {ms}$

But I realized that it doesn't take account for the $mgsin30$ force acting down the slope. Including that gives an answer of ≈ $6.075\,$m/s.

Any help would be appreciated! (Taking account of how simple the question is)


There are four forces acting on $P$:

  1. Friction

  2. $0.25$ N force up the plane

  3. Gravity

  4. Normal force

but only the first three contribute non-zero terms to the energy equation, since the normal force acts at right angles to $P$'s velocity. So you expect the right-hand side of the energy equation to have three terms.

The energy gained by $P$ due to the action of gravity is in the third line:

$0.2g \times 3 \sin 30$ Joules

You can either think of this as a force $0.2g\sin 30$ N acting over a distance of $3$ meters along the slope, or as a force $0.2g$ N acting over a vertical distance of $3 \sin 30$ meters, which is where the $mgh$ expression comes from. But if you add a fourth term to the right-hand side of the energy equation then you are double counting the contribution due to gravity.


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