Conservation of energy for a system

[_buddha]

Hello everyone I am stuck on this question, hopefully someone can help me out, here is the question:

2kg block is sitting on surface while 3kg block is hanging.

So far this is what I am thinking, the 2 blocks acts as a system, initially the block is 1.5m high, so mgy gives me the initial potential energy and when that block falls the velocity gives me the kinetic energy. So the kinetic energy plus the thermal energy(friction) should equal my initial potential energy. So this is what I have ..

mgy = 1/2mv^2 + fkd
73.5 = 36.1 + fkd
Fkd = 37.4 J

Which is wrong, the answer is -8.0 J.

 

[PhanthomJay]

Your PE term is incorrect. The block on the table has no change in PE. Only the falling block has a change in PE. Also, your fkd is on the wrong side of the equation, which affects the plus or minus sign of the solution.

 

[iRaid]

Your PE term is incorrect. The block on the table has no change in PE. Only the falling block has a change in PE. Also, your fkd is on the wrong side of the equation, which affects the plus or minus sign of the solution.

Why wouldn't there be any PE on the block on the table? Isn't there gravitational potential energy? (Just for my own knowledge, thanks)

 

[PhanthomJay]

Well, yes, if you take the reference axis for PE as the bottom of the 1.5m fall, it does have initial PE of 2g(1.5). But its final PE is also 2g(1.5). So you'd have to include that value on the right side of the equation, which you did not. The change in its PE is zero. So your PE term on the left for both blocks is correct...but on the right, you must add in the PE term for the block on the table.

 

[iRaid]

Well, yes, if you take the reference axis for PE as the bottom of the 1.5m fall, it does have initial PE of 2g(1.5). But its final PE is also 2g(1.5). So you'd have to include that value on the right side of the equation, which you did not. The change in its PE is zero. So your PE term on the left for both blocks is correct...but on the right, you must add in the PE term for the block on the table.

That's what I thought, btw, I'm not OP I was just curious as to what you meant :p

 

[PhanthomJay]

Oh, yeah, not the OP...:uhh:

 

[_buddha]

thanks for the help, actually that guy isn't me LOL. anyways i understand what youre saying, however i don't get why fkd should be on the left side? What i have now is U1 + U2 = K + U2 + fkd. The U2's cancel out...and the left hand side is my initial Emech(both U1 and U2 prior to the drop). isn't it thermal energy fkd PLUS the kinetic energy of both masses = to when it was both stationary?

 

[PhanthomJay]

thanks for the help, actually that guy isn't me LOL. anyways i understand what youre saying, however i don't get why fkd should be on the left side? What i have now is U1 + U2 = K + U2 + fkd. The U2's cancel out...and the left hand side is my initial Emech(both U1 and U2 prior to the drop). isn't it thermal energy fkd PLUS the kinetic energy of both masses = to when it was both stationary?

Ok, just to be sure we are on the same page, your equation now reads

3g(1.5) + 2g(1.5) = 1/2 (5)(3.8)² + 2g(1.5) + fkd, from which, fkd = +8. Correct? But the answer is fkd = -8. Why? Because the problem is asking for the work done by friction, whereas you have come up with the mechanical energy lost due to friction and transformed to thermal/sound/other forms of energy.

 

[_buddha]

ooh ok, makes sense! ty