In the figure below, Block A has mass 4.63 kg and Block B has mass 4.83 kg. Bloc
ID: 2097275 • Letter: I
Question
In the figure below, Block A has mass 4.63 kg and Block B has mass 4.83 kg. Block A slides on the horizontal table with coefficient of kinetic friction 0.424. The pulley spins with the taut string, so that the string does not slip as it passes over the pulley (what is the relationship between string speed and tangential speed of the outer surface of the pulley?). Treat the pulley as a solid disc with mass 0.555 kg and radius 1.32 cm. What is the speed of Block B after it has fallen by 1.19 m? Assume that both blocks are initially at rest and that the pulley spins without losing energy to friction. Write your answer in SI units to three significant figures.
Explanation / Answer
frictional force ,f =mg = 0.424 x 4.63 x 9.8 = 19.238 N
let tension in the string be T
so, net force on A & B are:
T-f = mA a
and, mBg - T = mB a
so, a = mBg - f/( mA + mB ) = 2.970 m/s2
1) so, velocity of Block B, after it has fallen 1.9m :
v= (2ah) = 7.630 m/s
2)
also, if the speed of the string and the velocity of block has to be the same , as the string is taut or else it would slack.
so. the tangential speed of the pulley must be equal to the speed of the string ,so as to prevent slipping .
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