How would you approach this problem: You have a frictionless pulleyon the edge o

Question

How would you approach this problem: You have a frictionless pulleyon the edge of a table,over which runs a cord to which are attacheda mass of 5 kg (which is hanging off the table initially) and atthe other end of the cord, a mass of 3 kg initially resting on thetable. Frictional coeff. for 3 kg mass/table is 0.4 ; what is thespeed of the 5 kg mass after it has fallen for 1.5 m? I was thinking of getting the frictional force, then the pos.acceleration from it, then subtract gravitational 9.8 m/s^2 from itto get net acceleration, the plug into eqn vf^2 = vi^2 + 2 g deltay But probably one has to use potential and kinetic energies forthis one. What say you? I was thinking of getting the frictional force, then the pos.acceleration from it, then subtract gravitational 9.8 m/s^2 from itto get net acceleration, the plug into eqn vf^2 = vi^2 + 2 g deltay But probably one has to use potential and kinetic energies forthis one. What say you?

Explanation / Answer

for block of mass 5 kg : ------------------------- tension in the cable mg - T = ma                                T = mg- ma                                T = 49 - 5a    ----( 1) where m = 5 kg For block of 3 kg : ------------------ T - f = m' a where f = frictional force = m ' g = 0.4 * 3* 9.8 = 11.76 N So, T = 11.76 + 3 a     ---( 2) from eq ( 1 ) and ( 2 ) , 49 - 5a = 11.76 + 3a       8 a = -11.76+49          a =4.655 m / s 6 2 Initial velocity u = 0 distance S = 1.5 m from the relation v ^ 2- u ^ 2 = 2aS speed after fall of 1.5 m is v = [ 2aS]                                            = 3.736 m /s

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