Speed and coefficeint of friction part 2

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The discussion revolves around a physics homework problem involving three stones that explode and move apart on a frozen lake. The key points include the need to calculate the total energy released in the explosion, given that 20% of this energy converts to kinetic energy. Participants emphasize using conservation of momentum to find the speed of stone C and then calculate the total kinetic energy of all stones. There is confusion about the relevance of the coefficient of friction provided in the problem, as it seems unnecessary for solving the kinetic energy and momentum aspects. Ultimately, the calculation for total kinetic energy after the explosion is confirmed to be correct, leading to a total energy release of approximately 88.3 J.
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Homework Statement



Three smooth stones, A, B and C are initially at rest, in contact with each other on the smooth surface of a frozen lake. The masses of the stones are A = 400 g, B = 300 g and C = 120 g and the coefficient of sliding friction between the stones and the ice is μslide = 0.0150. An explosion between the stones causes them to fly apart across the surface of the ice. Stone A flies off due North with an initial speed of 4.50 m s−1. Stone B flies off due east with an initial speed of 6.60 m s−1.

If 20% of the total energy of the explosion was converted to the kinetic
energy of the stones, what was the total energy released in the explosion?

Homework Equations



KE= 1/2 * M * V^2

PE= mgh

The Attempt at a Solution



total energy= ke+ pe

work done= force*distance

force= uR

μslide = 0.0150

i don't know R. And i don't know the distance travelend to calculate work done.

Need help please.
 
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1. Use cons of momentum to find the momentum and then the KE of stone C.
2. The KE of A + B + C equals 1/5 of the energy in the explosion.
You don't use PE, work done, or frliction.
 
why have they given friction in the question at all??
i didnt use friction in the first part either.
 
imy786 said:
why have they given friction in the question at all??
i didnt use friction in the first part either.
I don't see a need for it either. (here)
 
imy786 said:
why have they given friction in the question at all??
i didnt use friction in the first part either.
Maybe it is a smokescreen to make you realize that friction plays no part in collisions.
 
stone C= m1u1 + m2u2 = m1v1 +m2v2

m1= 0.120 kg
u1= 0

m2= would it stone B + stonce c mass
u2=? i got no idea what value to put here, as i have the speed of stone b and c.

could i not just calculate the total energy by just calculating KE , or do i have to also calculate conversation of momentum?
 
imy786 said:
stone C= m1u1 + m2u2 = m1v1 +m2v2

m1= 0.120 kg
u1= 0

m2= would it stone B + stonce c mass
u2=? i got no idea what value to put here, as i have the speed of stone b and c.

could i not just calculate the total energy by just calculating KE , or do i have to also calculate conversation of momentum?
Your first equation is not relevant here.
Start with momentum conservation
0=m_A{\vec v_A}+m_B{\vec v_B}+m_C{\vec v_C}.
to find v_c. Then use energy.
 
https://www.physicsforums.com/showthread.php?t=161727

speed of c is aprox 36meters/sec in the south west direction.

KE after collison= 0.5mava^2 + 0.5Mbvb^2 + 0.5McVc^2
= 0.5 ( 0.4 * 4.5^2 + 0.3*6.6^2 + 0.120* 36^2)
= 0.5 (8.1+ 13.068+ 155.52)
=0.5* 176.688
=88.3 J
 
is this correct to for the calculation to work out KE
 
  • #10
imy786 said:
https://www.physicsforums.com/showthread.php?t=161727

speed of c is aprox 36meters/sec in the south west direction.

KE after collison= 0.5mava^2 + 0.5Mbvb^2 + 0.5McVc^2
= 0.5 ( 0.4 * 4.5^2 + 0.3*6.6^2 + 0.120* 36^2)
= 0.5 (8.1+ 13.068+ 155.52)
=0.5* 176.688
=88.3 J
Are you sure that's the value you found for the speed of stone C? Looking at the other thread, I think you determined you had made a mistake with that. Double check it.
 
  • #11
is the method correct though to work it calculating KE
 
  • #12
It's right for finding the total kinetic energy after the explosion.
 
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