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=  `\vec {V_i}` = 0 m s⁻¹  

Final Velocity of Electron=  `\vec {V_f}` = 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 = `\frac {1}{2}` m `\v_f²` - `\frac {1}{2}` m `\v_i²`

Or [ as `\v_i` = 0 m s⁻¹] So


Fｘd = `\frac {1}{2}` m `\v_f²` 


by putting the corresponding values

Fｘ0.05 m =  `\frac {1}{2}` 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 = `\frac {18.2 ｘ10⁻¹⁷ kg m² s⁻²}{0.05 m}`

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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