Calculating Distance Travelled Using Electric Potential

[Jake357]

Homework Statement

Two small bodies have a mass of 0.009 kg each and are tied together with a wire that has a length of 2m. The bodies have the same electric charge of 5*10^-6 C each. The charges are on a horizontal surface. The wire gets cut. Calculate the distance travelled by each body till they stopped and the maximum velocities that they had. Calculate the distance between them at the moment they were having the maximum velocity. The coefficient of friction between the bodies and the surface is 0.1.
The the distance travelled by each body till they stopped must be 5.25 m.
The maximum velocity that they had must be 0.67 m/s.
The distance between them while they are having the maximum velocity must be 5 m.

Relevant Equations

Wf (the work of friction)=μmgd
We (work of the electric potential)=kq^2/d

I only could calculate the distance travelled by each body, by making the difference between the initial and final electric potential work equal to the work of friction done by the 2 bodies.

 

[haruspex]

What can you say about the forces on a body when it reaches maximum velocity?

 

[Jake357]

What can you say about the forces on a body when it reaches maximum velocity?

The only forces acting on the bodies are: electric, frictional and gravitational. Gravitational is in the y direction, so it doesn't affect the x direction forces, which makes it not useful. The only useful forces are frictional and electric. Their direction is opposite, because the electric force makes them repel each other, but the frictional one wants them to stop so it's in the opposite direction.
And also the electric force is dependent on the distance between them, which means that at a certain distance between them they will have maximum velocity and a different electric force.

 

[haruspex]

The only forces acting on the bodies are: electric, frictional and gravitational. Gravitational is in the y direction, so it doesn't affect the x direction forces, which makes it not useful. The only useful forces are frictional and electric. Their direction is opposite, because the electric force makes them repel each other, but the frictional one wants them to stop so it's in the opposite direction.
And also the electric force is dependent on the distance between them, which means that at a certain distance between them they will have maximum velocity and a different electric force.

Right, but specifically what can you say about those two horizontal forces when the object is at maximum velocity?

 

[kuruman]

To @Jake357 : Here is a hint to what @haruspex is trying to convey. A quantity, in this case the speed, is at a maximum at a point where it stops increasing and starts decreasing. What does that imply about the forces acting on the object when there are only two of them?

 

[Jake357]

To @Jake357 : Here is a hint to what @haruspex is trying to convey. A quantity, in this case the speed, is at a maximum at a point where it stops increasing and starts decreasing. What does that imply about the forces acting on the object when there are only two of them?

They are equal to each other.

 

[Jake357]

Right, but specifically what can you say about those two horizontal forces when the object is at maximum velocity?

The acceleration is zero, which means that the forces are equal to each other.

 

[haruspex]

The acceleration is zero, which means that the forces are equal to each other.

Right.

 

[Jake357]

Right.

kq^2/(2x+l)^2=mgμ
(2x+l)^2=kq^2/(mgμ)=25
2x+l=5
x=5-l/2=1.5 m

kq^2/l=2mv^2/2+kq^2/(2x+l)+2mgμx
mv^2=kq^2/l-kq^2/(2x+l)-2mgμx=0.0405 J
v^2=0.0405/m=4.5
v=2.12 m/s which is wrong. The velocity must be 0.67 m/s.

 

[haruspex]

kq^2/(2x+l)^2=mgμ
(2x+l)^2=kq^2/(mgμ)=25
2x+l=5
x=5-l/2=1.5 m

kq^2/l=2mv^2/2+kq^2/(2x+l)+2mgμx
mv^2=kq^2/l-kq^2/(2x+l)-2mgμx=0.0405 J
v^2=0.0405/m=4.5
v=2.12 m/s which is wrong. The velocity must be 0.67 m/s.

I agree with your working and answer.
Looks like the book answer made a factor of 10 error in calculating $v^2$.