Question #c1aad

1 Answer
May 16, 2016

5beta-4alpha5β4α

Explanation:

Assuming the spring obeys Hooke's law:
F=kxF=kx

Where
FF=force
kk=spring constant
xx= extension

The question implies that alphaα and betaβ is the total length of the spring, rather than the extension. If we define the length of the spring without any weights on it as yy, then

We know that
4=k(alpha-y)4=k(αy) .......equation (1)
and
5=k(beta-y)5=k(βy) .......equation (2)

If we divide the two equations:
5/4=(beta-y)/(alpha-y)54=βyαy
We can expand and solve for y

5(alpha-y)=4(beta-y)5(αy)=4(βy)
5alpha-5y=4beta-4y5α5y=4β4y
5alpha-4beta=y5α4β=y ..........equation (3)

For the 9N weight, of total length, say zz:
9=k(z-y)9=k(zy)

If we divide this by equation (2)

9/5=(z-y)/(beta-y)95=zyβy

9beta-9y=5z-5y9β9y=5z5y
9beta-4y=5z9β4y=5z
Substitute in for y from equation (3):

9beta-4(5alpha-4beta)=5z9β4(5α4β)=5z
9beta-20alpha+16beta=5z9β20α+16β=5z
25beta-20alpha=5z25β20α=5z
z=5beta-4alphaz=5β4α