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Question:

Suppose, a chemical bomb is stationary. Then it explodes into n parts in all direction. What is work done?

My Efforts

The bomb is stationary, so momentum and and kinetic energy is 0.

After explosion total momentum $\overrightarrow{P} $ is 0. So $\overrightarrow{P_1} + \overrightarrow{P_2} + ... + \overrightarrow{P_n}=0$

Kinetic energy after explosion $K = 1/2 mv^2$

So, change in Kinetic energy is $\Delta K= K$

According to Work Energy Theorem $W = \Delta K = K$ ($W$ is work done)

But, according to formula $W = F \times d = 0$

So, question is What is work done?

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    $\begingroup$ What is your question exactly ? Since you know the formula for Work-kinetic energy theorem, work is simply $W=\Delta K$, where $\Delta K$ is the total increase in the kinetic energy of the system. Why do you feel $\vec{F}\cdot\vec{d}$ is zero ? $\endgroup$ – Gaurav Nov 8 '14 at 6:43
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    $\begingroup$ Are you asking which 'force' is doing the work ? $\endgroup$ – Gaurav Nov 8 '14 at 6:46
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    $\begingroup$ @Simha because external force is 0 so work done is 0. I am asking whether W=K or W=0. $\endgroup$ – Freddy Nov 8 '14 at 6:56
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    $\begingroup$ I does not matter whether an internal force or external force is acting. It's just that if the force changes the kinetic energy of an object, it has done 'work'. $\endgroup$ – Gaurav Nov 8 '14 at 7:04
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    $\begingroup$ @Simha I can't gae you, can you please explain in detail as answer. $\endgroup$ – Freddy Nov 8 '14 at 7:13
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The work done was expended in moving all the components of the bomb away from it, largely dissipated as heat.

Picture the process in microscopic slow motion:

Chemicals transform from a solid to a gas + heat (which will increase the gas pressure some more). This isn't really "work" in the classical mechanics form (yet).

Gas pressure will exceed the limits of it's container (otherwise we wouldn't call it a bomb), container fractures absorbing some heat and kinetic energy in the process. We can call this work.

The container pieces are given considerable acceleration by the high-pressure gas, we call the pieces shrapnel at this point. The center of mass does not move and the sum of all the acceleration vectors is zero, but if you calculate each one individually and add up the forces without the vector you will get a rather large number. This is where most of the work will go.

The gas continues to expand, often as a shock wave, cooling rapidly in accordance to Charles' law. The energy in the gas is dissipated either into accelerating anything it hits or simply dumped into the atmosphere.

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if we really see the scenario carefully......the work-kinetic energy theorem is applicable only when there are no non conservative forces involved in the process... IN MORE GENERAL SENSE;(1)'WORK' IS DEFINED AS THE CHANGE IN ENERGY OR (2) WE MAY CALL IT TRANSFER OF ENERGY FROM ONE PART OF A SYSTEM (ISOLATED SYSTEM.......MEANS NO EXTERNAL FORCES ACT ON THE SYSTEM) TO THE OTHER PART IN WHICH CASE THE WORK IS DONE BY TRANSFER OF ONE KIND OF ENERGY TO THE OTHER FORM BY THE INTERNAL FORCES INVOLVED. In the case of explosion of a bomb, we can apply the relation between work and change in kinetic energy in the following form-

$W_{net}= ∆K + ∆H$ where $W_{net} = W_{int(conservative)} +W_{ext}$ and $∆H$ is the change in heat energy of the system.

The above quoted statement is essentially the law of conservation of energy applied to the system. IT SHOULD BE NOTED THAT THE WORK-KINETIC ENERGY THEOREM IS A SPECIAL CASE OF THE PRINCIPLE OF CONSERVATION OF TOTAL ENERGY.(THIS SPECIAL CASE IS- THE SYSTEM IS ISOLATED SO $W_{ext} = 0$ AND THE FORCES ACTING WITHIN THE SYSTEM ARE ONLY CONSERVATIVE.)

In case anyone wants reference use HALLIDAY RESNICK

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    $\begingroup$ Note that using all capitals as you have makes it appear you are telling at us. Consider using italics (surround the words with an asterisk) or bold (surround words with two asterisks) or block quote with the greater than symbol. $\endgroup$ – Kyle Kanos Nov 27 '14 at 13:50
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Work done by a force in a system does not depend on whether the force is internal or external. In your case, the internal forces are doing work. The net work done by these internal forces is $W=\Delta K =K$, where '$K$' is the final kinetic energy of the system.

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