A 2.2 kg piece of wood slides on the surface shown in the figure. Help, please!
ID: 3896262 • Letter: A
Question
A 2.2 kg piece of wood slides on the surface shown in the figure. Help, please!
A 2.2kg piece of wood slides on the surface shown in the figure below. The curved sides are perfectly smooth, but the rough horizontal bottom is 8m long and has a kinetic friction coefficient of 0.28 with the wood. The piece of wood starts from rest h = 4.0 m above the rough bottom. Find the distance along the rough patch from position A to where the wood will eventually come to rest. Will the block cross the rough patch at least once ? if so how high will it rise on the other side after the first pass?Explanation / Answer
a)
let u is the velocity at the bottom,
u = sqrt(2*g*h) = sqrt(2*9.8*4) = 8.854 m/s
workdone by friction, w = -mue*m*g*d
= -0.28*2.2*9.8*8
= - 48.2944
let V is velocity after crossing the rough surface.
we know, workdone = chnage in kinetic enrgy
W = 0.5*m*(v^2-u^2)
v = sqrt(u^2 + 2*W/m)
= sqrt(8.854^2 - 2*48.2944/2.2)
= 5.873 m/s
when it comes back, it will come to stop on rough surface after travelling x distance.
0.5*m*v^2 = mue*m*g*x
==> x = 0.5*v^2/mue*g
= 0.5*5.873^2/0.28*9.8
= 6.285 m
so, at 8 - 6.285 = 1.7125 m from point A the body comes to rest.
b) h = v^2/2*g
= 5.873^2/(2*9.8)
= 1.76 m
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