Multiple-object systems without friction: A series of weights connected by very light cords are given an upward acceleration of 4.00 m/s² by a pull P, as shown in the figure. A, B, and C are the tensions in the connecting cords. The SMALLEST of the three tensions, A, B, and C, is closest to A) 80.0 N. B) 196 N. C) 276 N. D) 483 N. E) 621 N.

Question

Quizplus · Accepted Answer

First, we need to find the acceleration of each weight. Since there is no friction, all weights will have the same acceleration. Using Newton's second law, we have:

P - A - B - C = m*a

where m is the total mass of the system (m = m1 + m2 + m3 = 3 kg) and a is the acceleration.

We know that a = 4.00 m/s^2 and we can solve for P:

P = m*a + A + B + C = 3*4.00 + A + B + C = 12 + A + B + C

Next, we need to find the tension in each cord. We can look at each weight separately and apply Newton's second law again. For example, for the top weight, we have:

Tension in cord A - m1*g = m1*a

where g is the acceleration due to gravity (g = 9.81 m/s^2) and m1 is the mass of the top weight (m1 = 1 kg). Solving for tension in cord A, we get:

Tension in cord A = m1*(g + a) = 1*(9.81 + 4.00) = 13.81 N

Similarly, for the middle weight, we have:

Tension in cord B - m2*g = m2*a

where m2 is the mass of the middle weight (m2 = 1.5 kg). Solving for tension in cord B, we get:

Tension in cord B = m2*(g + a) = 1.5*(9.81 + 4.00) = 19.215 N

Finally, for the bottom weight, we have:

Tension in cord C - m3*g = m3*a

where m3 is the mass of the bottom weight (m3 = 0.5 kg). Solving for tension in cord C, we get:

Tension in cord C = m3*(g + a) = 0.5*(9.81 + 4.00) = 6.905 N

Now, we need to find the smallest tension. It is clear that tension in cord C is the smallest, and it is equal to 6.905 N. Therefore, the answer is C.

Multiple-Object Systems Without Friction: a Series of Weights Connected by Very