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## Homework Statement

Block A and block B are on a frictionless table as shown in Figure 2. Spring 1 connects block A to a frictionless peg at 0 and spring 2 connects block A and block B. Their masses are respectively m

_{A}=0:45 kg and m

_{B}=0:32 kg. When the blocks are in uniform circular motion about 0, the springs have lengths of l

_{A}=0.80 m and l

_{B}=0.50 m. The springs are ideal and massless, and the linear speed of block B is 2.8 m/s. If the distance that spring 2 stretches is 0.070 m, determine the spring constant of spring 2.

## Homework Equations

F=ma

a=v

^{2}/r

F=kx

## The Attempt at a Solution

y-direction is perpendicular to surface and x-direction is parallel to surface

r

_{A}=l

_{A}, r

_{B}=l

_{A}+l

_{B}

T

_{1}=Tension is spring 1, T

_{2}=Tension in spring 2

N

_{A}=Normal acting on block A, N

_{B}=Normal acting on block B

x

_{B}=0.07m (The extension of spring 2)

For block A (From FBD)

Resolving (y-direction): F

_{Ay}=m

_{A}a

_{Ay}=N

_{A}-m

_{A}g=0

Resolving (x-direction): F

_{Ax}=m

_{A}a

_{Ax}=T

_{1}-T

_{2}

⇒(m

_{A}v

_{A}

^{2})/r

_{A}=T

_{1}-T

_{2}

For block B (From FBD)

Resolving (y-direction): F

_{By}=m

_{B}a

_{By}=N

_{B}-m

_{B}g=0

Resolving (x-direction): F

_{Bx}=m

_{B}a

_{Bx}=T

_{2}

⇒(m

_{B}v

_{B}

^{2})/r

_{B}=T

_{2}

And this is where I have become stuck, because I know I can't simply say T

_{2}=kx

_{B}as this would give the spring constant for a single spring that has extended to a total length of 1.3m. And continuations from this point do not seem to go anywhere e.g. substituting centripetal force of block B into the equation for block A, does not help as far as I can see.