Two blocks are connected by a string that passes over a frictionless pulley, as

Question

Two blocks are connected by a string that passes over a frictionless pulley, as shown in the figure. The pulley has a mass of mp = 2.00 kg, and can be treated as a uniform solid disk that rotates about its center. Block A, with a mass mA = 3.00 kg, rests on a ramp measuring 3.0 m vertically and 4.0 m horizontally. Block B hangs vertically below the pulley. Note that you can solve this exercise entirely using forces, torques, and the constant-acceleration equations, but see if you can apply energy ideas instead. Use g = 10 m/s2. When the system is released from rest, block A accelerates up the slope and block B accelerates straight down. When block B has fallen through a height h = 2.0 m, its speed is v = 4.00 m/s.

(a) Assuming that no friction is acting on block A, what is the mass of block B?

(b) If, instead, there is friction acting on block A, with a coefficient of kinetic friction of 5/8, what is the mass of block B?

Explanation / Answer

a) angle of the incline = @ = tan^-1(3/4) = 37 deg

lets assume initial PE = 0

initial KE = 0

fnally :

PE of block B = mg(-2) = -2mg

PE of block A = 3g(2sin37) = 3.61g

KE of B = m(4^2)/2 = 8m

KE of A = 3 x (4^2)/2 = 24 J

KE of pulley = Iw^2 /2 = (mr^2 /2 ) (v/r)^2 /2 = mv^2 / 4 = 2 x (4)^2 /4 = 8 J

Using energy conservation,

initial energy = final energy

0 = -2mg + 3.61g + 8m + 24 + 8   

11.62m = 67.41

m = 5.80 kg

B) On A perpendicular to the incline,

N = mgcos37

friction = uN = (5/8) x 3 x 9.81 x cos37 = 14.67 J

work done by friction = - 14.67 x 2 = - 29.38 J

work done by friction + work done by gravity = change in KE

-29.38 + 2mg - 3.61g = 24 + 8m + 8 - 0

m = 9.34 kg

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