How much work is required to stop an electron

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SUMMARY

The work required to stop an electron with a mass of 9.11×10-31 kg moving at a speed of 1.40×106 m/s is calculated using the formula for kinetic energy (KE = ½mv²). The correct kinetic energy value is 8.93×10-19 J. While the work done to stop the electron is technically negative, the problem only requires the magnitude, which is 8.93×10-19 J, without the negative sign. The confusion arises from the interpretation of the sign in the context of the problem.

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Homework Statement



How much work is required to stop an electron (m=9.11×10−31kg) which is moving with a speed of 1.40×10^6 m/s ?

2. Homework Equations

The Attempt at a Solution



E.g a particle has KE=10J. The work that must be done on it to stop it is -10J.

So work out the kinetic energy of the electron(½mv²) and put a minus sign in front. I get -8.93x10⁻¹⁹ J.but when i answer the question it says i am wrong? can someone please help me
 
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No need of minus sign dear.
Minus sign represents the direction. But question don's ask a direction. So remove the minus sign. Then it'll be okay.:smile:
 
rperez1 said:
E.g a particle has KE=10J. The work that must be done on it to stop it is -10J.
Technically, you are correct: The work done must be negative. But apparently they just want the magnitude of the work, not the sign.
 
Doc Al said:
Technically, you are correct: The work done must be negative. But apparently they just want the magnitude of the work, not the sign.
it still says incorrect when i take out the negative which makes no sense bc i have no other way to get another answer.
 
Jimmy Moriaty said:
No need of minus sign dear.
Minus sign represents the direction. But question don's ask a direction. So remove the minus sign. Then it'll be okay.:smile:
it still continues to say that it is incorrect, which makes no sense bc i know i did it correctly, i have no other way to get another answer. i don't know what is wrong with my answer.
 
Do they specify the units they want you to use? (Joules makes sense, but there are other units.)
 

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