Two blocks are connected by a string that passes over a massless, frictionless p
ID: 1347234 • Letter: T
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 = 2.00 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 heighth = 2.0 m, its speed is v = 6.00 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 inclinstaion, theta = tan^-1(3/4)
= 36.87 degrees
let a is the acceleration of the blocks
a = (v^2 - u^2)/(2*h)
a = (6^2 - 0^2)/(2*2)
= 9 m/s^2
Let T is the tension in the string.
Net force acting on A, FnetA = T - mA*g*sin(36.87)
mA*a = T - mA*g*sin(36.87)
T = mA*a + mA*g*sin(36.87)
= 2*9 + 2*9.8*sin(36.87)
= 29.76 N
net force acting on blockB, FnetB = mB*g - T
mB*a = mB*g - T
T = mB*(g - a)
mB = T/(g-a)
= 29.76/(9.8 - 9)
= 37.2 kg
b)
Let T is the tension in the string.
Net force acting on A, FnetA = T - mA*g*sin(36.87) - mue_k*m*g*cos(36.87)
mA*a = T - mA*g*sin(36.87) - mue_k*m*g*cos(36.87)
T = mA*a + mA*g*sin(36.87) + mue_k*m*g*cos(36.87)
= 2*9 + 2*9.8*sin(36.87) + (5/8)*2*9.8*cos(36.87)
= 39.56 N
net force acting on blockB, FnetB = mB*g - T
mB*a = mB*g - T
T = mB*(g - a)
mB = T/(g-a)
= 39.56/(9.8 - 9)
= 49.45 kg
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