A particle is placed on a rough plane which is inclined to the horizontal at an angle θ, where tanθ =4/3, and released from rest. The coefficient of friction between the particle and the plane is 1/3. Find the particle's acceleration.

Start by drawing a diagram with all of the information from the question on it, as well as the forces acting on the particle. As forces can be divided into their component parts, we can look at the question in terms of forces acting perpendicular to and parallel to the slope. Resolving perpendicular to the plane, R=mg cosθ (eq. 1). Resolving parallel to the plane, resultant force = ma --> mg sinθ - F = ma (eq. 2). As the particle is going to move down the plane, the friction between the particle and the plane will be its maximum, hence F=μR. Putting this into eq. 2 --> mg sinθ - μR = ma. Putting eq. 1 into this new equation --> mg sinθ - μmg cosθ = ma --> g sinθ - μg cosθ = a. To get the decimal values of the trig functions, either use your calculator or draw a right angled triangle. From the triangle, chose one angle to be θ then put the length of the "opposite" side as 4 and the length of the adjacent side as 3. The hypotenuse of this triangle is therefore 5 (Pythagoras' theorem) , and from this triangle you can work out sinθ = 4/5, and cosθ=3/5. Now change all of the symbols for values that you know (where g = 9.8) --> 9.8*0.8 - (1/3)9.80.6 = a = 5.88m/s²