Instead of having Fx=-kx where x is the amount stretched and the spring is unstr
ID: 2144752 • Letter: I
Question
Instead of having Fx=-kx where x is the amount stretched and the spring is unstretched at x=0, the actual force given by the string is approximately given by Fx = -kx for x is less than S and Fx = -k2(x-S)-k1S for x is greater than or equal to S Here k1 and k2 are known constants
a)Given the force suppose a block of mass m starts at x=0 with a velocity of v1 to right of a frictionless table. Determine how far it would go before turning around if v1 is so small is doesn't go beyond x=S.
b)Given the force suppose a block of mass m starts at x=0 with a velocity of v1 to right of a frictionless table. Determine how far it would go before turning around if v1 is large enough to go beyond x=S.
I have the answer key, so please work out the problem and just show the work.
Explanation / Answer
a) PE = - integral of F dx = 1/2 kx^2
so 1/2 m v1^2 = 1/2 kx^2
x = sqrt( m v1^2/k)
b) now PE = - integral of Fdx from 0 to S - integral of F dx from S to x
= 1/2 kS^2 + integral of k2 x - k2 S + k1 S from S to X
= 1/2 kS^2 + 1/2 k2 x^2 - 1/2 k2 S^2 - k2 S (x-S) + k1 S ( x-S)
so
1/2 mv^2 = 1/2 kS^2 + 1/2 k2 x^2 - 1/2 k2 S^2 - k2 S (x-S) + k1 S ( x-S)
solve for x