In the figure below, a small block is sent through point A with a speed of 9.6 m
ID: 2229808 • Letter: I
Question
In the figure below, a small block is sent through pointAwith a speed of9.6m/s. Its path is without friction until it reaches the section of lengthL= 12 m, where the coefficient of kinetic friction is 0.70. The indicated heights areh1=7.4m andh2=2.3m.
(a) What is the speed of the block at pointB?
(b) What is the speed of the block at pointC?
(c) Does the block reach pointD?
If so, what is its speed there and if not, how far through the section of friction does it travel?
Explanation / Answer
a) Use (K + U)A = KB So 1/2*m*vB^2 = 1/2*m*vA^2 +m*g*h1 so vB = sqrt(vA^2 +2*g*h1) = sqrt(8.7^2 + 2*9.8*7.6) = 15.0m/s b) KB = (K + U)C So KC = KB - UC...1/2*m*vC^2 = 1/2*m*vB^2 - m*g*h2 So vC = sqrt(vB^2 -2*g*h2) = sqrt(15^2 - 2*9.8*1.8) = 13.8m/s c) If the work done by friction is > KC then the block does not reach D W = µ*m*g*L = 0.70*g*L = 0.70*9.8*12 = 82.3J/kg KC = 1/2*v^2 = 1/2*13.8^2 = 95.2J/kg so the block reaches D with a speed of 1/2*m*vD^2 = 1/2*m*vC^2 - µ*m*g*L so vD = sqrt(vC^2 - 2*µ*g*L) = sqrt(13.8^2 -2*0.70*9.8*12) = 5.08m/s