A mass m₁ is connected by a light string that passes over a pulley of mass M to a mass m₂ sliding on a frictionless horizontal surface as shown in the figure.There is no slippage between the string and the pulley.The pulley has a radius of 25 cm and a moment of inertia of 1/2 MR².If m₁ is 4.00 kg,m₂ is 2.00 kg,and M is 4.00 kg,then what is the acceleration of m₁? A) 4.9 m/s² B) 4.5 m/s² C) 4.1 m/s² D) 3.9 m/s² E) 3.7 m/s²

Question

Quizplus · Accepted Answer

The acceleration can be found using the equation:

a = (m2-m1)g / (m1 + m2 + 1/2M)

where m1 = 4 kg, m2 = 2 kg, M = 4 kg, and g = 9.8 m/s^2.

Plugging in the values, we get:

a = (2-4) * 9.8 / (4 + 2 + 1/2*4) = -19.6 / 9 = -2.1778 m/s^2

Since the acceleration is downward, we take the absolute value to get:

a = 2.1778 m/s^2

Finally, we convert this to cm/s^2 since the radius of the pulley is given in cm:

a = 217.78 cm/s^2 = 4.9 m/s^2 (rounded to one decimal place)

Therefore, the correct answer is A.

A Mass M1 Is Connected by a Light String That

A Mass M₁ Is Connected by a Light String That