Two blocks are connected by a string that passes over a massless, frictionless p

Question

Two blocks are connected by a string that passes over a massless, frictionless pulley, as shown in the figure. Block A, with a mass mA = 4.00 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 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?
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

tantheta = 3/4

theta = 37 degress

for mass A

T - mA*g*sintheta = mA*a........(1)

for mass B

mB*g - T = mB*a.........(2)

1 + 2

mB*g - mA*g*sintheta = a*(mA+mB)

acceleration a = g*(mB - mA*sintheta)/(mA+mB)

speed after h

v = sqrt(2*a*h)

given v = 4 m/s

4 = sqrt(2*(mB-4*sin37)*9.8*2/(4+mB))

mB = 6.83 kg

(b)

for mass A

T - mA*g*sintheta - uk*mA*g*costheta = mA*a........(1)

for mass B

mB*g - T = mB*a.........(2)

1 + 2

mB*g - mA*g*sintheta - uk*mA*g*costheta = a*(mA+mB)

acceleration a = g*(mB - mA*sintheta - uk*mA*costheta)/(mA+mB)

speed after h

v = sqrt(2*a*h)

given v = 4 m/s

4 = sqrt(2*(mB-4*sin37-(5/8)*4*cos37)*9.8*2/(4+mB))

mB = 10.2 kg

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