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## Homework Statement

A particle of mass 4kg is being towed at a constant speed up a rough plane inclined at 30 degrees to the horizontal by a force 4g N acting parallel to the slope. At the top of the slope the particle moves onto a rough horizontal slope with the same coefficient of friction. If the towing force continues to act in the same direction, show that the particle undergoes an acceleration of

[tex]\frac{g\sqrt{3}}{6}[/tex] m.second square

## Homework Equations

basically use F=ma and get the answer where m=4kg

## The Attempt at a Solution

1. On the inclined plane coefficient of friction was calculated as 1/ (sqrt 3)

2.Since they say the force of 4g will be acting in the same direction even when it is on the horizontal plane i took this too mean that 4g will be acting at an angle 30 degrees to the horizontal when the particle is on a horizontal plane..

This way Reaction force would be 4g-4gsin30= 2g

And net force would be equal too

4a= 4gcos30 - 2g/ (sqrt 3)

which makes a = [tex]\frac{g\sqrt{3}}{3}[/tex] m.second square which is twice the answer that is given.

Can some one please check this out. Its from an A level Mechanics book