Abox can slide with negligible friction on a horizontal table top. a string of negligible mass runs horizontally over a dowel with negligible friction, and then connects to an identical box which is falling vertically. m m m~g how does the tension in the string between the boxes compare to the net force on the boxes? 1. it is less, about 50%.

it is less, about 50%. Since we're ignoring friction, let's do the math and see what happens. I'll use the variables: M = mass of a single box. g = gravitational acceleration. The only force available is from the box that's suspended, so we'll have (1) Mg Newtons of force available. Now since the 2 boxes are connected together, that force will have to accelerate 2M. So the acceleration of the group is gM/2M = g/2 So both boxes are accelerating at half the local gravitational acceleration. This makes sense since the gravitational acceleration is only affecting half of the mass being accelerated. Now, to see the tension in the string, you need to ask the question, "How much force is required to accelerate the mass sliding horizontally at g/2?" The answer to that question is obviously Mg/2 newtons. So the net force being supplied is Mg newtons (See equation (1) above), and the force being transmitted via the string is Mg/2 Newtons, it's pretty obvious that the tension in the string is 50% of the total net force.