How Does Spring Displacement Affect Block Motion in a Loop-the-Loop?

[joemama69]

Homework Statement

A small block is launched on the fricntionless loop the loop shown. te sprink launcher has a sprink constant k

a)find the velocity at the top of the loop as a function of the displacement x of the spring launcher from its equilibrium length

b)find the minimum cvalude of x such that the block goes over the top in contact with the track

c)find the normal force on the bock A as a function of x

Homework Equations

The Attempt at a Solution

I believe i would have to use .5ks², and then calculate the acceleration, and aply it to uniform circular motion, is this correct

W = .5kx²

K = .5mv²

v = (2W/m)^.5 = (kx²/m)^.5

 

[joemama69]

ok pretend my first attemp never happened

Part A
T_A = .5ks²
U_A = 0

T_B = .5mv²
U_B = -2Rmg

.5ks² = .5mv² - 2Rmg

v = ((ks² + 4Rmg)/m)^.5

Part B

same as above, just solved for s

s = ((mv²-4Rmg)/k)^.5

Part C

not really sure about the Normal Force, Does it still point up even if its on the top of the loop the loop

 

[rl.bhat]

When the block is moving in the loop, two forces are acting. One centripetal force and the other the weight of the block. Mg*cosθ contributes to the normal reaction.

 

[joemama69]

so the normal is

N = mgcos(180)

do u agree with my other responses, they do not match what the answer is said to be

Part B answer is xmin = (5mgR/k)1/2

Part C answer is N(x) = (kx2/R) - 5mg

where does that 5 mg come from