A block is released from rest at the top of a plane inclined at an angle of 54 d

Question

A block is released from rest at the top of a plane inclined at an angle of 54 degrees. The coefficient of kinetic friction varies along the plane according to the relation mu k = sigma x, where x is the distance along the plane measured in meteres from th etop and where sigma=.6 m^-1. The acceleration of gravity is 9.8 m/s^2. Determine how far the block slides before coming to rest. (Hint: Find the acceleration of the block and integrate to find the velocity. Note that (dv/dt) = (dv/dx)(dx/dt) Answer in units of m.

Explanation / Answer

horizontal component of speed is 4.9 * cos(30) = 4.24m/s vertical component of speed is 4.9 * sin(30) = .5m/s d = 1/2at^2, where d = height of table = 2.0m a = 9.8m/s^2, so t = sqrt(2(2)/9.8) = 0.639s until it hits the floor d = vt = 4.24*0.639 = 2.7m from the table

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