In the figure below, a small block is sent through point A with a speed of 9.7 m

Question

In the figure below, a small block is sent through point A with a speed of 9.7 m/s. Its path is without friction until it reaches the section of length L = 12 m, where the coefficient of kinetic friction is 0.70. The indicated heights are h1 = 7.3 m and h2 = 2.4 m.

(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

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