In the figure a block slides along a path that is without friction until the blo

Question

In the figure a block slides along a path that is without friction until the block reaches the section of length L = 0.62 m, which begins at height h = 1.2 m, on a ramp of angle theta = 31 degree. In that section the coefficient of kinetic friction is 0.410 The block passes through point A with a speed of 8.8 m/s. If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A? Number _____ Unit _____ the tolerance is +/-2%

Explanation / Answer

Let's assume that it does make it to point B. Call the mass of the block (M)
Let's start off by assuming that it is frictionless all the way.
Point B is at a vertical height above A of (h + Lsin(31°)) = 1.52 m

When M reaches point B it has converted Mg * 1.52 of its initial kinetic energy into g.p.e
So without friction its k.e. at point B would have been ½M *8.8² - Mg*1.52 = 23.8M joules {g = 9.81 m/s²}

How much of that energy is actually lost as work done against friction?
OK, this is going to involve the relationship between the coefficient of friction (), the friction force (Ff) and the normal force (Fn)

Ff = * Fn
Ff = 0.41 * Mg.cos(31°) = 3.45M newtons
Work done against friction = Ff * L = (3.45M * 0.62) = 2.139M joules

So the actual k.e. at point B will be: (23.8M - 2.139M) = 21.66M joules

k.e. = ½Mv² = 21.66M
v² = 43.32
v = 6.58 m/s

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