Two roofers are installing a new roof. Bob is sitting on the top of the roof that is angled at 23 degrees from the horizontal. Nate is sitting 5 m further down the roof. In order to get a box of shingles to Nate, Bob just slides the box directly down to him. If the coefficient of kinetic friction for the box and roof is 0.8, then with what speed should Bob initially push the box so that it slides to a stop right as it reaches the Nate?
- NCSLv 71 month ago
So, what are you studying right now, kinematics or work/energy?
the acceleration down the slope is
a = g*(sinΘ - µ*cosΘ)
a = 9.8m/s² * (sin23º - 0.8*cos23º) = -3.39 m/s²
(meaning the acceleration is actually UP the slope!)
v² = u² + 2ad
0² = u² - 2 * 3.39m/s² * 5m = 5.82 m/s
initial energy - friction work = final energy = 0 J
(at rest and at reference frame 0 elevation)
½mv² + mgh - µmgd*cosΘ = 0
multiply by 2/m
v² + 2gh - 2µgd*cosΘ = 0
and plug in your knowns:
v² + 2*9.8m/s²*5m*sin23º - 2*0.8*9.8m/s²*5m*cos23º = 0
v² = 33.9 m²/s²
v = 5.82 m/s
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- oubaasLv 71 month ago
conservation of energy shall appy :
m/2*V^2+m*g*5*sin 23 = m*g*cos 23*μ*5
mass m cancels :
V = √ 2*5*9.806*(0.921*0.8-0.391) = 5.8 m/sec