<h2>Given :</h2>
<h2>Solution :</h2>

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Answer:
Explanation:
Given
Electric Field 

velocity 
mass of electron 
Force on a charge Particle moving in Magnetic Field
![a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Be%5Cleft%20%5B%20%5Cvec%7BE%7D%2B%5Cleft%20%28%20%5Cvec%7Bv%7D%5Ctimes%20%5Cvec%7BB%7D%5Cright%20%29%5Cright%20%5D%7D%7Bm%7D)

That would be kinetic energy
Answer:
x' = 1.01 m
Explanation:
given,
mass suspended on the spring, m = 0.40 Kg
stretches to distance, x = 10 cm = 0. 1 m
now,
we know
m g = k x
where k is spring constant
0.4 x 9.8 = k x 0.1
k = 39.2 N/m
now, when second mass is attached to the spring work is equal to 20 J
work done by the spring is equal to


x'² = 1.0204
x' = 1.01 m
hence, the spring is stretched to 1.01 m from the second mass.