# Finding speed in the system

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.