Two blocks are connected by a string that passes over a frictionless pulley, as
ID: 1349439 • Letter: T
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?
PLEASE REMEMBER TO USE G = 10! NOT G = 9.8!
Explanation / Answer
Apply v^2-u^2 = 2*a*S
4^2 - 0 = 2*a*2
16 = 4*a
a = 4 m/s^2
net torque is T = I*alpha
Let T1 be the tension in the inclined string
T1-(mA*g*sin(theta)) = mA*a = 3*4= 12
T1 = 12+(3*10*3/5) = 30 N
and mB*g -T2 = mB*a
T2 = mB*(g-a) = mB*(10-4) = 6*mB
(T1-T2)*R = 0.5*mp*R^2(a/R)
T1-T2 = 0.5*mp*a = 0.5*2*4 = 4
30-6mB = 4
6mB = 26
mb = 26/6 = 4.33 kg
-------------------------------------------------------------------------
T1-(mA*g*sin(theta))-(mu_k*mA*g*cos(theta)) = mA*a = 3*4= 12
T1 - (3*10*3/5) - ((5/8)*3*10*(4/5)) = 12
T1 = 45 N
then T1-T2 = 0.5*mp*a = 0.5*2*4 = 4
T2 = T1-4 = 45-4 = 41 N
6*mB = 41
mB = 41/6 = 6.83 kg
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.