posted by Gage on .
I THINK SOMEONE HAS THIS SAME QUETION! A massless spring of constant k = 91.0 N/m is fixed on the left side of a level track. A block of mass m = 0.50 kg is pressed against the spring and compresses it a distance of d. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.5 m. The entire track and the loop-the-loop are frictionless, except for the section of track between points A and B. Given that the coefficient of kinetic friction between the block and the track along AB is µk = 0.24, and that the length of AB is 2.5 m, determine the minimum compression d of the spring that enables the block to just make it through the loop-the-loop at point C. [Hint: The force of the track on the block will be zero if the block barely makes it through the-loop-the-loop.]
Assuming point C is at the top, then
g= v^2/r so you have to calculate v at the top.
So energy at the bottom must equal KE at the top + the change in PE to get to the top
RequiredKEbotom= 1/2 m*vtop^2+mg(2r)
Now, you have losses before you get there due to friction, mu*mg*2.5, so that has to be added to get the initial KE at the spring.
PEspring= mu*mg*2.5+1/2m vtop^2+mg(2r)
PE spring= 1/2 k d^2, solve for d.
thank you, but i am exteremly confused! so i just use PE spring to find d?
i guess i'm just confused on the mu means static friction? .24?
is mu friction? .24?