Two masses are suspended by cord that passes over a pulley with negligible mass.The cord also has negligible mass.One of the masses,m₁,has a mass of 5.0 kg and the other mass,m₂,has a mass of 3.0 kg.The acceleration of m₁ is A) 3.3 m/s² upward. B) 3.3 m/s² downward. C) 2.5 m/s² upward. D) 2.5 m/s² downward. E) 1.1 m/s² upward.

Question

Quizplus · Accepted Answer

Since only one of the masses has acceleration, we can use the formula:

F = ma

where F is the net force on the system, m is the mass of the system, and a is the acceleration of the system.

In this case, the net force on the system is the difference between the weights of the two masses:

Fnet = mg1 - mg2

where g is the acceleration due to gravity (9.8 m/s^2), g1 is the weight of m1 (49 N), and g2 is the weight of m2 (29.4 N).

Fnet = 49 N - 29.4 N = 19.6 N

Now we can plug this into the formula:

Fnet = ma

19.6 N = (5 kg)(a)

a = 3.92 m/s^2

This is the acceleration of m1. Since the acceleration is downward, we know that the force of gravity on m1 is greater than the tension in the cord, so m1 is accelerating downward.

Now we can use the formula for the tension in the cord:

T = m2g - m2a

T = (3 kg)(9.8 m/s^2) - (3 kg)(3.92 m/s^2)

T = 29.4 N - 11.76 N = 17.64 N

This is the tension in the cord. Since m2 is accelerating upward, the tension in the cord is also upward.

Therefore, the answer is (D) 2.5 m/s^2 downward.

Two Masses Are Suspended by Cord That Passes Over a Pulley