Speed and coefficeint of friction part 2

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Homework Help Overview

The problem involves three stones on a smooth surface, initially at rest, that are propelled apart by an explosion. The masses of the stones and the coefficient of friction are provided, along with the initial speeds of two stones after the explosion. The goal is to determine the total energy released in the explosion based on the kinetic energy of the stones.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using conservation of momentum to find the kinetic energy of the third stone, C, and question the relevance of friction in the context of the problem.
  • Some participants express uncertainty about the need for potential energy and work done in their calculations.
  • There is a suggestion to calculate total energy solely from kinetic energy, while others emphasize the importance of momentum conservation.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the necessity of certain parameters, such as friction. Some guidance has been offered regarding the use of conservation of momentum and kinetic energy calculations, but no consensus has been reached.

Contextual Notes

Participants note confusion regarding the role of friction in the problem, suggesting it may not be relevant to the calculations needed for the explosion's energy. There is also mention of potential mistakes in previous calculations that may affect the discussion.

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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
[tex]0=m_A{\vec v_A}+m_B{\vec v_B}+m_C{\vec v_C}.[/tex]
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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