Question

The particle p is released from rest at the top of a smooth plane inclined at angle a where sina=16/65.the distance travelled by P from top to bottom is S metres and speed of P at bottom is 8ms^-1 find S and hence speed of P when it has travelled 1/2S?

Answer 1

`s=13.27" m"`
`v=5.66" m s"^-1`

Explanation:

Draw a diagram!!!!

I've drawn the only forces acting on P; its weight (`mg`) and the reaction force (`R`). Since we are not doing work against any other forces ie friction (R is perpendicular to the acceleration so doesn't count) we can use the principle of conservation of energy.

`"change in GPE"="change in KE"`

As a reminder, `E_g=mgh` and `E_k=1/2mv^2`

To first find GPE, we need h (marked on my diagram as `h_0`). We can find this using trigonometry:

`h_0=ssinalpha`
`=16/65s`

`:. "GPE"=mgh_0`
`=16/65mgs`

Yes, m is an unknown, so we need to get rid of it at some point.

`DeltaE_k=1/2mv^2`
`=1/2mxx8^2`
`=32m`

by conservation of energy:

`32cancelm =16/65cancelm gs`

`gs=130`

`s=130/g`
Taking `g=9.8` (you may need to take it to `9.81` depending on your course)

`s=650/49`
`s=13.27"m (4sf")`

---

We can do the same thing with `1/2s` to work out `v` (I forgot to put v on the diagram).

`h_1=1/2ssinalpha`
`=1/2xx650/49xx16/65`
`=80/49`
`=1.633m`

`"GPE"=mgh_1`
`=80/49mg`

`"KE"=1/2mv^2`

We don't know m or v; this is as simple as we can go right now.

By conservation of energy

`80/49mg=1/2mv^2`

`80/49xx9.8=1/2v^2`

`32=v^2`

`v>0` o we can square root both sides

`v=4sqrt2`
`=5.66"m s"^-1`

Answer 2

Alternate solution.

Explanation:

Particle p of mass `m` moves from rest at the top of a smooth inclined plane of angle `a` under action of gravity. Using Newton's Second Law of motion

`vecF=mveca`

Acceleration in the downwards direction is equal to `sin a` component of the gravity.

`a=(mgsina)/m`
`=>a=gsina`
`=>a=9.81xx16/65 \ms^-2`

Distance `S` is found from the kinematic expression

`v^2-u^2=2as`

Inserting given values we get

`8^2-0^2=2(9.81xx16/65)S`
`=>S=64/(2(9.81xx16/65))`
`=>S=13.25\ m`

Using same kinematic expression to find velocity when particle traveled a distance of `1/2S`

`(v_(S/2))^2+0^2=2(9.81xx16/65)8/2`
`=>v_(S/2)=sqrt(8(9.81xx16/65))`
`=>v_(S/2)=4.4\ ms^-1`