Question

Three boxes slide on a frictionless horizontal surface when pulled by a force of magnitude F. When we compare the tensions T1 and T2 with the force F, we find that

1) T1 = T2 = F.
2) T1 = F > T2.
3) F > T1 = T2.
4) F > T1 > T2.
5) F - T1 < T1 - T2.

Answer 1

The correct relation between the force F and the tensions T1 and T2 while pulling three boxes on a frictionless surface in a line is F > T1 > T2, as each tension must be less than the force before it to account for the force exerted on the box in front.

Step-by-step explanation:

The question concerns three boxes sliding on a frictionless horizontal surface being pulled by a force F. Without friction, the only forces to consider are the applied force F and the tensions T1 and T2 in the ropes connecting the boxes. Assuming the boxes are sliding at a constant speed, the net force on each box is zero according to Newton's First Law of Motion, which is also known as the law of inertia.

If the boxes are arranged in a line and being pulled from one end, the force F is applied on the first box. The tension in the rope connecting the first and second box (T1) must be less than F because some of the force is used to accelerate or maintain the motion of the first box. Similarly, the tension in the rope between the second and third box (T2) must be less than T1 for the same reason. Therefore, the correct option in this case would be: F > T1 > T2. This reflects the fact that each successive tension must be smaller to account for the force exerted on each preceding box.

Answer 2

The tension in the string between the boxes is the same on either side of each box because the surface is frictionless, so T1=T2=T3=F.

let's consider the three boxes individually.

If they are connected by a string, and the string is being pulled by a force F, the tension in the string will vary along its length.

Let's label the boxes as A, B, and C from left to right. The force F is applied to box A.

The tension in the string is denoted as T1, T2, and T3 between boxes A and B, B and C, and C and the external force F, respectively.

The tension in the string is transmitted through each box. Since the surface is frictionless, the tension in the string is the same on either side of each box.

Now, let's analyze the forces acting on each box:

1. Box A:
   • Tension T1 acts to the right.
   • External force F acts to the right.
   • There are no other horizontal forces.
   • Net force on box A is F, so T1 = F.
2. Box B:
   • Tension T1 acts to the left.
   • Tension T2 acts to the right.
   • There are no other horizontal forces.
   • Net force on box B is zero, so T1 = T2.
3. Box C:
   • Tension T2 acts to the left.
   • Tension T3 acts to the right.
   • There are no other horizontal forces.
   • Net force on box C is zero, so T2 = T3.

Considering all the boxes together:

• T1 = F (from box A).
• T1 = T2 (from box B).
• T2 = T3 (from box C).

So, T1 = T2 = T3 = F.

Therefore, the correct answer is 1) T1 = T2 = F.