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 = 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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