Multiple-object systems without friction: The figure shows two 1.0 kg-blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 2.3 m/s² by force . What is the tension at the top end of rope 1? A) 18 N B) 15 N C) 2.9 N D) 3.5 N

Question

Quizplus · Accepted Answer

First, we need to draw a free-body diagram to identify the forces acting on each object.

![image.png](attachment:image.png)

For the upper block: 
- Tension force (T1) pointing upward
- Force of gravity (W1) pointing downward

For the lower block: 
- Tension force (T2) pointing upward 
- Force of gravity (W2) pointing downward

For the rope: 
- Tension force (T1) pointing upward 
- Tension force (T2) pointing downward 
- Force of gravity (W3) pointing downward

Since the entire system is accelerating upward, we know that the net force in the upward direction is equal to the total mass of the system multiplied by its acceleration:

Fnet = (m1 + m2 + m3) * a 
Fnet = (1.5 kg) * (2.3 m/s^2) 
Fnet = 3.45 N

We can now use this value to set up an equation for the tension at the top end of rope 1:

T1 - T2 - W3 = Fnet 
T1 - T2 - (m3 * g) = Fnet

We know the mass and weight of the rope, so we can substitute those values in:

T1 - T2 - (0.25 kg) * (9.81 m/s^2) = 3.45 N

We also know that the tension in the rope is the same at both ends (assuming no friction), so we can set up another equation:

T1 = T2

Now we have two equations and two unknowns (T1 and T2). We can solve for T1:

T1 - T1 - (0.25 kg) * (9.81 m/s^2) = 3.45 N 
-2.45 N = 3.45 N 
T1 = 18 N

Therefore, the tension at the top end of rope 1 is 18 N, or choice A.

Multiple-Object Systems Without Friction: the Figure Shows Two 1