KE Theorem Help

Question

a 27.5 kg block is being pulled by a force F = 262 N up along a rough incline. The coefficient of kinetic friction between the block and the incline is μk = 0.40. The block is pulled through a distace of 28.0 m.

The block starts from rest at the bottom of the incline. After it is pulled through a distance of 28.0 m up along the incline, what is its vertical height (in meters), hf?

I found the work by all the non conservative forces to be 4816.65 but I'm not really sure what to do next.

Any help would be appreciated.

Answer 1

Draw the Free Body Diagram of the problem and find the various components of the forces involved!!

Known quantities:

Applied force, F = 262.0N

mass of block, m = 27.5 kg

displacement, d = 28.0m

coefficient of kinetic friction, μ_k = 0.40

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The applied Force, F = 262.0N = mgsinθ,

where g = 9.81m/s² is acceleration due to gravity, and θ is the inclined angle (with the horizontal)

From this equation, solve for the angle, θ

=== > θ = sin^-1(262.0/mg) ≈ 76°

With this angle and the given displacement, we can find the vertical height, h using

h = dsinθ = (28m)(sin76) = 27.2 m

This is what you are looking for. However, you can go a step further by invoking the energy theorem and then solve for the non conservative forces. Here is how to go about it

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From KE Theorem

W = PE + KE + E_lost, where W = total work done, PE = potential energy, KE = kinetic energy and E_lost = energy lost from conservative forces, example, friction.

Recall that PE = mgh = (27.5)(9.81)(27.2) = 7337.9 J

Friction, F_k = μ_kN = μ_kmgcosθ = 0.4(27.5)(9.81)(cos76) = 26.1 N

Therefore the lost energy, due to friction will be

E_lost = -F_kd = - 731 J

I hope this helps :-)