Question

Problem 1: In each experiment below. two blocks start from rest at time t₁ on a level, frictionless surface; both hands exert the same magnitude force F_o over the same time interval from  t₁ to  t₂, and the blocks do not run into each other during this interval. In each case, consider the system of the two blocks plus, if applicable, any objects connecting the two blocks to each other. Rank the five experiments by the net external work done on each system.

Problem 1 (Explanation Task) In each experiment below, two blocks start from rest at time t¡ on a level, frictionless surface; both hands exert the same magnitude force F, over the same time interval from tį to t2, and the blocks do not run into each other during this interval. In each case, consider the system of the two blocks plus, if applicable, any objects connecting the two blocks to each other. Rank the five experiments by the net external work done on each system. Experiment 1 m m Experiment 2 m m Experiment 3 m Experiment 4 m m/2 Experiment 5 Rigid rod т m

Please answer using the following structure:

Answer 1

Work done is given as dot product of the force and displacement.

W = F.ds

Experiment 1:

Assuming the right direction to be positive we have force on left block as F and right block as -F.

When it comes to displacement as both blocks are of same mass and opposite force is applied for fixed interval of time thus the displacement of both masses will be opposite in direction but of same magnitude.

So, the work done by the system will be W = F.d + -F.-d = 2F_o.d; d = F_o/m x (t₂-t₁)²/2

=> W = F_o²(t₂ - t₁)²/m

Experiment 2:

Assuming the right direction to be positive we have force on left block as F and right block as F.

When it comes to displacement as both blocks are of same mass and the same force is applied for fixed interval of time thus the displacement of both masses will be same.

So, the work done by the system will be W = F.d + F.d = 2F_o.d; d = F_o/m x (t₂-t₁)²/2

=> W = F_o²(t₂ - t₁)²/m

Experiment 3:

Here assume the spring force to be k,

Now if we see F_net = F_o - 2kx;

The displacement will be much less than that when it was displaced without having spring so, if we assume it to be L. Then the net work done will be W = 2 x (F_oL - kL²)

Although L is very small Still the work done will become negative; as the negative terms is L squared and that it is multiplied to k which is often a large value sometimes. Thus here mostly W<0;

Experiment 4:

Assuming the right direction to be positive we have force on left block as F and right block as -F.

When it comes to displacement as both blocks are of different mass and force applied is in opposite direction for fixed interval of time thus the displacement of both masses will be opposite in direction and of different magnitude.

So, the work done by the system will be W = F.d₁ + -F.-d₂;  d₁ = F_o/m x (t₂-t₁)²/2; d₂ = -2F_o/m x (t₂ - t₁)²/2

=> W = F_o²(t₂ - t₁)²/2m + -F_o/2 x -F_o/m x (t₂ - t₁)² = F_o²(t₂ - t₁)²/m

Experiment 5:

Since the blocks here are connected to a rigid rod when the blocks are applied force the blocks will try to move against each other but they couldn't move as the rod will not allow any displacement that will forbid its rigidity. You can consider it in force method also, Now since it is a rod it will apply a force T which will be same throughout the rod. Now assume that net acceleration of left mass is a, which will be given as

a_L = (F_o - T)/m

Now the right mass will also experience same forces but in opposite direction so,

a_R = -(F_o - T)/m

Now as we can see that the acceleration of any mass will be dependent on the point on the rod connecting the mass. Thus if mass moves the point on rod will also move. which means it will also have the same acceleration as we calculated above. Now since the rod is rigid the acceleration of left point has to be same as the right point.

a_L = a_R

=> T = F_o

Which means that acceleration of both masses have to be zero.

Thus no work is done here on the blocks. W = 0

Ranking:- (Positive comes first then negative)

1) Exp:- 1, 2 and 4 (same by calculation)

2) Exp:- 5 (Zero)

3) Exp:- 3 (negative by observation)