One side of the roof of a building slopes up at 33.5°. A roofer kicks a round, flat rock that has been thrown?
One side of the roof of a building slopes up at 33.5°. A roofer kicks a round, flat rock that has been thrown onto the roof by a neighborhood child. The rock slides straight up the incline with an initial speed of 15.0 m/s. The coefficient of kinetic friction between the rock and the roof is 0.430. The rock slides 10.0 m up the roof to its peak. It crosses the ridge and goes into free fall, following a parabolic trajectory above the far side of the roof, with negligible air resistance. Determine the maximum height the rock reaches above the point where it was kicked.
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Let the mass of the rock = m.
Force between rock and roof due to rock’s weight = F = mg.cosϑ = mg.cos 33.5º
Work done in friction sliding up the roof = Wr = Fμ.10 where μ (= 0.43) is the friction coefficient.
Wr = 10.μ.m.g.cos33.5º
The rock’s initial energy as it leaves the roofer’s foot = m.15²/2
So rock’s kinetic energy as it leaves the top of the roof = KE = m.(15²/2 - 10.μ.g.cos33.5º - 10g.sin33.5º)
Note that the rock loses kinetic energy due to climbing up the roof, not just to friction. This is represented in the third term in the brackets of the last expression.
Its velocity = V = √(2.KE/m) = √[2.(15²/2 - 10.μ.g.cos33.5º - 10gsin33.5º)]
The vertical upward component of this velocity = V’ = V.sin33.5º = √[2.(15²/2 - 10.μ.g.cos33.5º - 10g.sin33.5º)].sin33.5º
This upward component V’ will be subject to a deceleration = g and will reach zero at a time t after leaving the roof where t = V’/g
The height the rock reaches above the point where it leaves the roof is then = h = ½gt²
= (15²/2 - 10.μ.g.cos33.5º - 10gsin33.5º).sin²33.5º
So the height the rock reaches from point it was kicked = (15²/2 - 10.μ.g.cos33.5º - 10 sin33.5º).sin²33.5º + 10.sin33.5º
Just plug in the numbers to get the numerical result (g = 9.81 m/s²).
The final equation is wrong but very close.
Part 1:The final Velocity found using the conservation of energy equation ∆K+∆U+∆Eint is right,
however when trying to find the height from the top of the roof to the maximum height of the disk is wrong.
Part 2:(total energy of the system)=∆K+∆U
(Kf-Ki)+(Uf-Ui)
(0-(1/2mv^2))+(mgh-0)
Getting rid of the mass/changing the velocity into x-axis/solving for height you get:
((v•sinØ)^2)÷(2g)=H
*notice the velocity is only changed into x-axis on the second part because the first part of the equation has an altered x/y axis to simplify the equation*
The total height is:
∆H=( H+ (10sinØ))
For those of you that are trying this problem at 11:55 P.m. here is the right equation
[[√(2•(15^2)/2)-(10•(Uk)•g•cos(Ø))-(10•g•sinØ))|•( sinØ)^2)÷(2g)] +10sinØ]