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 = 1.00 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 = 3.00 m/s.

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

(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?
___ kg

Explanation / Answer

from the figure tantheta = 3/4

theta = 37 degrees

initial potential energy of the system Ei = 0

B comes down by distance h and A moves up the incline by a distance h

final potential energy of the system Ef = -mB*g*h + mA*g*h*sintheta + (1/2)*mA*v^2 + (1/2)*mB*v^2 + (1/2)*I*w^2

I = moment of inertia of pulley = (1/2)*mp*r^2

r = radius of pulley

w = angular speed = v/r

Ef = -mB*g*h + mA*g*h*sintheta + (1/2)*mA*v^2 + (1/2)*mB*v^2 + (1/2)*mp*v^2

Ef = -mB*g*h + mA*g*h*sintheta + (1/2)*v^2(mA + mB + mp)

from energy conservation

Ei = Ef

0 = -(mB*9.8*2) + (1*9.8*2*sin37) + (1/2)*3^2*(1 + mB + 2)

mB = 1.66 kg <<<<-------ANSWER

=====================================

part (b)

work done by friction Wf = uk*mB*g*H*costheta

Ei - Ef = Wf

Ei = Ef + Wf

0 = -(mB*9.8*2) + (1*9.8*2*sin37) + (1/2)*3^2*(1 + mB + 2) + ((5/8)*1*9.8*2*cos37)

mB = 2.323 kg <<<<-------ANSWER

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