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1. A particle P is projected up a rough plane which is inclined at an angle of a to the horizontal where tan a=3/4. The coefficient of friction between the particle and the plane is 1/3. The particle is projected from the point A with speed 20 m/s an comes to instantaneous rest at point B.

Tan-1 (3/4) = 36.9

Show that while P is moving up the plane its deceleration is 13g/15.

I am completely lost on what to do.
2. Consider the particle P at a general point C in between A and B. Draw a force diagram, drawing the weight, the normal and the friction. Resolve parallel and perpendicular to the plane to find the particle accerelation up the plane.
Hint
Suppose that the normal reaction is and the deceleration is , and its mass is . Resolving perpendicular to the plane, we find that

, where

And resolving parallel to the plane, we find that:

, where is the coefficient of friction (given as 1/3).

Now solve to find .
3. R-10gcost 36.9
R= 10g cos 36.9

10gsin36.9 - 1/3 R =ma

Is that correct?
4. (Original post by Deep456)
R-10gcost 36.9
R= 10g cos 36.9

10gsin36.9 - 1/3 R =ma

Is that correct?
Remember that friction acts in the direction opposing motion so in your last equation, the sign on 1/3R is wrong.
5. (Original post by GHOSH-5)
Remember that friction acts in the direction opposing motion so in your last equation, the sign on 1/3R is wrong.
Distance AB = 23.5m(3sf)

time take from A>>>B =2.35s(3sf)

Find the speed of P when it returns to A, I have done the 2 bits in the middle, how do you do this last part?
6. (Original post by Deep456)
Distance AB = 23.5m(3sf)

time take from A>>>B =2.35s(3sf)

Find the speed of P when it returns to A, I have done the 2 bits in the middle, how do you do this last part?
You'll need to draw a whole new force diagram and find the new acceleration of the particle from B to A (which can be done in a similar fashion to before). Remember friction is now acting up the slope as the particle is moving down it. The use SUVAT from B to A to find it's speed at A.

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Updated: October 10, 2009
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