How large a force is required to accelerate an electron (m = 9.1 ｘ10⁻³¹ kg) from rest to a speed of 92 ｘ10⁷ m s⁻¹ through a distance of 5 cm. (Ans: 3.6 ｘ10⁻¹⁵ N)



Given:

Mass of Electron = m = 9.1 ｘ10⁻³¹ kg

Initial Velocity of Electron=  →Vi→Vi = 0 m s⁻¹  

Final Velocity of Electron=  →Vf→Vf = 2.0 ｘ10⁷ m s⁻¹

Distance traveled: d = 5 cm = 0.05 m

To Find:

Force =  F = ?   


Solution:


According to work energy principal

Work done = change in K.E.

Fｘd = 1212 m vf² - 12 m vi²

Or [ as vi = 0 m s⁻¹] So


Fｘd = 12 m vf² 


by putting the corresponding values

Fｘ0.05 m =  12 9.1 ｘ10⁻³¹ kg` ( 2.0 ｘ10⁷ m s⁻¹ )² 

Fｘ0.05 m =  4.55 ｘ10⁻³¹ kg ( 4 ｘ10¹⁴ m² s⁻²)

Fｘ0.05 m =  18.2 ｘ10⁻¹⁷ kg m² s⁻²

F = 18.2ｘ10⁻¹⁷kgm²s⁻²0.05m

F = 364 ｘ10⁻¹⁷ N

or

F = 3.64 ｘ10⁻¹⁵ N -------------Ans.

Hence the force required as 3.64 ｘ10⁻¹⁵ N




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