What are the key mistakes in approaching this work-energy principle problem?

[Radhika Ahuja]

Homework Statement

:
A skater with mass 70kg standing on ice throws a stone of mass 5kg with a velocity of 8m/s in the horizontal direction. Find the distance over which the skater will move back if the coeffcient of friction between the skates and the ice is 0.02[/B]

Homework Equations

The Attempt at a Solution

This is how I tried to solve it:

The force with which the stone if thrown = mass of stone x g = 5g
Thus, the reaction force the man experiences is -5g

Now, the net force in with which the person moves back would be (-5g + Force of friction)
And here friction is uN = 0.02x70g = 1.4g

=> Net force with which the person moves back is -5g + 1.4g = -3.6g

The Work done by this force is F.d = Change in Kinetic Energy/the energy that was transferred from the man to the stone

=> F.d = 1/2 x mass of stone x v^2
=> d = (2.5 x 64)/(3.6 x 9.8) ~ 4.53m

However, this answer is incorrect, the correct answer is 0.83m and the solution solves it using power.

I want to understand two things:
- why is this way of approaching it wrong/what am I missing?
- when I see a problem, how should I identify whether a problem will require using the Power equation/how should I approach problems like this in general[/B]

 

[Delta2]

I see two major problems with your approach:
1) The way you calculate the reaction force =-5g is wrong. What you calculate as 5g is the weight of the stone. I think we have insufficient data to calculate the reaction force.
2) The way you apply the work energy theorem. You set the work done on the skater from the net force on him, equal to the kinetic energy of the stone. That is wrong . The work of the net force on the skater is equal to the change in the kinetic energy of the skater.

My question is what is the Power equation that the problem uses? Also in the solution don't they use conservation of momentum in order to calculate the speed of the skater right after he throws the stone?

 

[Delta2]

Also I think the problem implies one critical assumption (I am not so sure about this but anyway I ll state it), that the throwing of stone is done within infinitesimally small time $dt$ so the reaction force is considered to be infinite (so that the impulse $Fdt=\infty\cdot 0$ of the reaction force is finite and equal to the momentum of the stone) and its of no further use to this problem.
Anyway can you post fully the solution that you read, I want to figure out some things about it.

 

[haruspex]

energy that was transferred from the man to the stone

The energy comes from the man's muscle, not from any KE the man had... and if it had come from the man's KE then that would have been negative work, slowing the man down.

 

[haruspex]

I think the problem implies one critical assumption (I am not so sure about this but anyway I ll state it), that the throwing of stone is done within infinitesimally small time dt

Technically that is true, because of the friction, but the effect would be minuscule.

 

[Radhika Ahuja]

I see two major problems with your approach:
1) The way you calculate the reaction force =-5g is wrong. What you calculate as 5g is the weight of the stone. I think we have insufficient data to calculate the reaction force.
2) The way you apply the work energy theorem. You set the work done on the skater from the net force on him, equal to the kinetic energy of the stone. That is wrong . The work of the net force on the skater is equal to the change in the kinetic energy of the skater.

My question is what is the Power equation that the problem uses? Also in the solution don't they use conservation of momentum in order to calculate the speed of the skater right after he throws the stone?

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You're right, they use conservation of momentum. Thank you!