A constant force produces maximum velocity V on the block connected to the spring of force constant K as shown in the fig. When the force constant of spring becomes 4K, then find maximum velocity of the block. Assume that initially the spring is in relaxed state.
Text Solution
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V/2
Sol. By work energy theorem;
w ext + w spring = k f – k i
Let x 1 , x 2 be the equilibrium distances of spring from natural length & V V’ are their velocities at equilibrium positions.
Initially,
Fx 1 – 
kx 1 2 = 
mv 2….. (1)
and finally Fx 2 – 
k'x 2 2 = 
mv' 2 …(2)
In both cases: force applied is same, and velocity becomes maximum when F = kx. (at equilibrium)
(after which the mass will deaccelerate)
∴ F = kx 1 = (4k)x 2
⇒ x 2 = 
(k´ = 4k)
Substituting in (2):
fx 2 – 
k´ x 2 2 = 
mv´ 2
– 
(4k) 
= 
mv´ 2
⇒ 
[Fx 1 – 
kx 1 2 ] = 
mv' 2 ….(3)
Dividing (3)/(1) ; we get :
= 
⇒ v ' = 
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