# Work problem

#### Swatch

A 8 kg block is being pulled horizontally on a table by a 6 kg block that hangs in the air by a cord fasten to the 8 kg block over a pulley, there is only friction between the table and the 8 kg block. Both blocks are initially moving with 0,900 m/s and they come to rest after 2 meters.
I have to use the work-energy theorem to find the coefficient of kinetic friction.

What I've done so far is:

total work done on the 6 kg block must be -0.5*6*0.9*0.9 = 2.43 J
So the net force must be -2.43J/2m = -1,215N
Fy=-mg+T=-1,215 so the tension is 57.58 N

Total work done on the 8 kg block must be -0.5*8*0.9*0.9 = -3.24 J
So the net force on the block must be
Fx=57.58-fk=-3.24J/2m
coefficient of friction must be 55.96/8*9.8=0.714

I don't get the right answer which is 0.786

Could someone please give me a hint to what I am doing wrong.

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#### Doc Al

Mentor
I suggest starting over and thinking in terms of the total change in energy of the system. What's the change in KE? What's the change in PE? The total change in energy must equal the work done by the friction.

#### Swatch

I have calculated the total kinetic energy change by using 8+6=14 as the mass and the speeds 0.9 m/s and 0 m/s That energy change is not enough to explain the friction. The book I'm reading has not mentioned potential energy so I am not supposed to use that to solve this. Sorry, I am not any closer to solving this.

#### geosonel

Swatch said:
I have calculated the total kinetic energy change by using 8+6=14 as the mass and the speeds 0.9 m/s and 0 m/s That energy change is not enough to explain the friction. The book I'm reading has not mentioned potential energy so I am not supposed to use that to solve this. Sorry, I am not any closer to solving this.

reason your approach is not working: you're only accounting for change in kinetic energy, when in fact potential energy is also changing:

friction work = ΔKE + ΔPE

you calculated ΔKE correctly.
you now need ΔPE

fyi, in this case:
ΔPE = mgΔH(8 kg block) + mgΔH(6 kg block)
where ΔH is height change (0 m for 8 kg block, and 2 m for 6 kg block).

you must account for ΔPE to get correct answer

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#### Swatch

O.K I got it right, including the PE, geosonel. But I don't get why I have to include this energy. I can see that mg*s for the 6 kilo block is the work that is done on that block, but I did already calculate the KE for it, or -0.5*m*v*v. So the friction does work to slow the 6 kilo block down and the 8 kilo. Please explain to me #### Doc Al

Mentor
Actually, your original method was fine. You just made an error in signs, confusing the direction of the forces.
Swatch said:
What I've done so far is:

total work done on the 6 kg block must be -0.5*6*0.9*0.9 = 2.43 J
So the net force must be -2.43J/2m = -1,215N
Fy=-mg+T=-1,215 so the tension is 57.58 N
Realize that the net force is upward. Calling up positive, your force equation should be: T - mg = +1.215. Recalculate the tension accordingly.

Total work done on the 8 kg block must be -0.5*8*0.9*0.9 = -3.24 J
So the net force on the block must be
Fx=57.58-fk=-3.24J/2m
coefficient of friction must be 55.96/8*9.8=0.714
Once you use the correct tension, you'll get the right answer.

If you had covered PE, then you could apply the "Work-Energy" theorem using total mechanical energy (KE + PE). I think that way is a bit easier, since the PE automatically takes care of gravity. But no matter, you can solve the problem just like you did, explicitly including the work done by gravity.

Another little trick is to realize that the work done by the tension must be zero. It does positive work on one block, negative work on the other: the total is zero.

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#### Swatch

Thank you, Doc Al. I understand this now.
Thanks guys.

#### OlderDan

Homework Helper
Doc Al said:
Another little trick is to realize that the work done by the tension must be zero. It does positive work on one block, negative work on the other: the total is zero. (Of course, if you realize that the tension is internal to the system, this will make sense. Only external forces can do net "work" on the system.)
I'm thinking your parenthetic statement is too broad. It would be true of internal forces of a rigid object, or a system with no relative motion between masses (effectively the case in this problem) but does not apply to some cases where there is relative motion, such as inelastic collisions. Sometimes the internal forces are conservative and can be related to a potential energy, but not always.

#### Doc Al

Mentor
OlderDan said:
I'm thinking your parenthetic statement is too broad. It would be true of internal forces of a rigid object, or a system with no relative motion between masses (effectively the case in this problem) but does not apply to some cases where there is relative motion, such as inelastic collisions. Sometimes the internal forces are conservative and can be related to a potential energy, but not always.
You are absolutely correct, Dan. Very sloppy of me. (I will remove that statement from my earlier post.) Thanks for the correction.

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