Abox sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest.
1)if the initial speed of the box were doubled, how far x2 would the spring compress?

x2 = ? 2 x1
x2 = 2 x1
x2 = 4 x1

x₂=2×1

Explanation:

Using work energy theorem, consider the gravitational potential energy at its lowest position of compression to be zero, because change in kinetic energy is 0 then;

mgx-(kx)²/2 =0   where m is mass of object, g is acceleration due to gravity, k is spring contant and x is the compression on the spring.

mgx=(kx)²/2

x=2mg/kcompression when velocity of object is Zero

But ΔK.E =-1/2mv²⇒kx²=mv² where v is velocity of the object and K.E is kinetic energy

So if kx²=mv² then

v=x *√(k/m) where v=0

Doubling v will multiply the x *√(k/m) by 2 making x₂ twice x₁

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach