Hello furor celtica Originally Posted by

**furor celtica** initially a small block of wood is at a point O on a rough plane inclined at 15° to the horizontal. the block is projected directly up the plane with initial speed 4ms^-1. The coefficient of friction between the block and the plane is 1/10. The block comes instantaneously to rest at A. Find the distance OA.

So I used v^2=u^2 + 2as to get s=8/a

then i used ma=µ(10mcos15) to get a but s turned up something like 8 and apparently thats wrong, can anyone show their working in this method?

You haven't included the component of the weight of block down the plane in your second equation. It should be:

$\displaystyle mg\sin\theta + F = ma$

where $\displaystyle \theta$ is the angle the plane makes with the horizontal, $\displaystyle \displaystyle F$ is the friction force down the plane and $\displaystyle \displaystyle a$ is the acceleration down the plane.

Resolving at right angles to the plane gives:

$\displaystyle N = mg\cos\theta$

where, of course, $\displaystyle \displaystyle N$ is the normal contact force between the block and the plane.

And then with

$\displaystyle F = \mu N$

you're there.

Grandad