MIT Physics Problem Quiz

What is the initial setup of the Atwood machine problem discussed in the video?

• A fixed pulley with a massless string and two blocks of different masses

• A fixed pulley with two blocks of equal mass

• A fixed pulley with a single block and a massless string

• A rotating pulley with two blocks of different masses

What must be true for the lighter block to move down the incline?

• The tension T1 must be less than the weight of the lighter block

• The tension T2 must be greater than the frictional force

• The angle Alpha must be greater than 45 degrees

• The coefficient of friction must be zero

Which equation represents the condition for the block to move up the incline?

• 2mg sin(alpha) - mu 2mg cos(alpha) > T1

• mg > 2mg sin(alpha) + mu 2mg cos(alpha)

• T1 > T2 > mg

• 2mg sin(alpha) + mu 2mg cos(alpha) < T2

What is the role of the moment of inertia in the equations of motion for the pulley?

• It affects the tension in the string

• It is used to calculate the angular acceleration of the pulley

• It determines the linear acceleration of the blocks

• It is irrelevant to the motion of the system

How are the accelerations of the blocks and the pulley related?

• The acceleration of the pulley is twice that of the blocks

• They are unrelated

• The acceleration of the blocks is half that of the pulley

• They are all equal

What is the expression for T1 derived in the video?

• T1 = mg/2 + T2/3

• T1 = mg/3 + 2T2/3

• T1 = 2mg/3 + T2/3

• T1 = mg + T2

What is the final expression for T2 in terms of the given variables?

• T2 = mg(4/7 + 6/7 sin(alpha) + 6/7 mu cos(alpha))

• T2 = mg(3/7 + 5/7 sin(alpha) + 5/7 mu cos(alpha))

• T2 = mg(2/7 + 4/7 sin(alpha) + 4/7 mu cos(alpha))

• T2 = mg(1/7 + 2/7 sin(alpha) + 2/7 mu cos(alpha))

What is the significance of the constraint that the string does not slip?

• It allows the use of rotational mechanics for the pulley

• It means the accelerations of the blocks and pulley are equal

• It simplifies the calculation of the moment of inertia

• It ensures that the tensions in the string are equal

How is the acceleration of the block calculated in the final section?

• By using the expressions for both T1 and T2

• By ignoring the frictional force

• By using the expression for T1 only

• By considering only the gravitational force

What is the final expression for the acceleration of the block?

• a = g(1 - 5/7 - 4/7 sin(alpha) - 4/7 mu cos(alpha))

• a = g(2/7 - 4/7 sin(alpha) - 4/7 mu cos(alpha))

• a = g(3/7 - 5/7 sin(alpha) - 5/7 mu cos(alpha))

• a = g(4/7 - 6/7 sin(alpha) - 6/7 mu cos(alpha))