Hi, just another question which im slighly unsure about but Ill see if you can help me.

A block of mass 20kg is being pulled along a rough horizonal surface by a rope inclined at an angle of 30degrees to the horizonal.

The coefficient of friction between the block and surface is μ. Model the block as a particle which slides on the surface.

(a) If the tension in the rope is 60 newtons, the block moves at a constant speed.

(i) Show that the magnitude of the normal reaction force acting on the block is 166N

I did $\displaystyle R+60cos60=20*9.8$ $\displaystyle R=196-60cos60$ $\displaystyle R=166N$ so I assume I got this bit correct.

(ii) Find μ

I did $\displaystyle F=60cos30$ $\displaystyle F=51.96$

F=μR 51.96=μx166 μ=$\displaystyle 51.96/166$ μ=0.131

(b) If the rope remains at the same angle and the block accelerates at $\displaystyle 0.8ms-2$, find the tension in the rope.

I said 'By Newtons 2nd Law Resultant Force = ma

so

$\displaystyle Tcos30-51.96=20*0.8$

$\displaystyle Tcos30=16+51.96$

$\displaystyle T=67.96/cos30$

$\displaystyle T=78.5N$

The last question im pretty sure I have no idea what Im talking about, and please excuse the maths talk, werent sure how to get all the symbols in correctly.

Thanks