Question Completion Status: QUESTION 2 Block B, with mass 5.00 kg, rests on bloc

Question

Question Completion Status: QUESTION 2 Block B, with mass 5.00 kg, rests on block A, with mass 8.00kg. There is no friction between the table and block A, but the coefficient of static friction between block A and B is 0.750. A "massless" string attached to block A passes over a frictionless pulley and block C is suspended from the other end of the string. If we want ALL the blocks moving together with the same acceleration, the largest possible mass that block C can have is 39.0 kg. Let's change the mass for block C and place a new block C with mass 50.0 kg instead. Evaluate if the new acceleration is supported by the friction force acting on block B. Determine: The total acceleration of the three blocks if they were to move all together. (This may not happen, just pretend.... 5.55 m/s2 9.42 m/s2 7.78 m/s2 0.79 m/s2

Explanation / Answer

Let, tension in the string = T

Net force on block C,

Fnet = M*g - T

From second law of motion,

M*a = M*g - T

T = M*g - M*a ......(1)

Net force on system of block A and B,

Fnet = T

T = (mA + mB)*a ........(2)

From eq(1) and (2),

M*g - M*a =  (mA + mB)*a

a = Mg / (mA + mB + M)

a = 50*9.8 / (5 + 8 + 50)

a = 7.78 m/s^2

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