Asked by Ana
                A massless spring of constant k = 88.9 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 = 0.34, 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.]
so i did,
(.5)*(88.9)*X^2=(0.34)*(.5)*(9.81)*(2.5)+(2)*(1.5)*(.5)*(9.81)
44.45^2=18.88425
X=.6517994474m
What seems to be my mistakes because hw site doesn't accept my answer.
            
            
        so i did,
(.5)*(88.9)*X^2=(0.34)*(.5)*(9.81)*(2.5)+(2)*(1.5)*(.5)*(9.81)
44.45^2=18.88425
X=.6517994474m
What seems to be my mistakes because hw site doesn't accept my answer.
Answers
                    Answered by
            Damon
            
    Potential energy stored in spring = (1/2)k d^2= 44.5 d^2
That will be the kinetic energy the block has departing the spring
KE =(1/2)mv^2 = 44.5 d^2
I assume that A to B is this next stretch between the spring and the bottom of the ferris wheelie.
normal force on track = m g
friction force = .34 m g
work done by friction = .34 m g (2.5)
= .85 m g
so kinetic energy at bottom of loop = 44.5 d^2 - .85 m g
Now
Loss of energy going up loop = m g h = 3 m g
Now at the top of the loop for zero force on track:
m v^2/r = m g
(1/2) m v^2 = m g r = (1/2)1.5 m g
Ke at top is therefore .75 m g
So the total kinetic energy at the bottom of the track must be 3.75 m g
so
3.75 m g = 44.5 d^2 - .85 m g
44.5 d^2 = 2.90 m g
d^2 = (2.90/44.5)(.5)(9.81)
d = .565
    
That will be the kinetic energy the block has departing the spring
KE =(1/2)mv^2 = 44.5 d^2
I assume that A to B is this next stretch between the spring and the bottom of the ferris wheelie.
normal force on track = m g
friction force = .34 m g
work done by friction = .34 m g (2.5)
= .85 m g
so kinetic energy at bottom of loop = 44.5 d^2 - .85 m g
Now
Loss of energy going up loop = m g h = 3 m g
Now at the top of the loop for zero force on track:
m v^2/r = m g
(1/2) m v^2 = m g r = (1/2)1.5 m g
Ke at top is therefore .75 m g
So the total kinetic energy at the bottom of the track must be 3.75 m g
so
3.75 m g = 44.5 d^2 - .85 m g
44.5 d^2 = 2.90 m g
d^2 = (2.90/44.5)(.5)(9.81)
d = .565
                    Answered by
            Ana
            
    again, I appreciate your work but site doesn't accept this answer. I don't know why this is occurring for all the problems that I'm working on.
    
                    Answered by
            Damon
            
    Beats me, but check my arithmetic carefully.
    
                    Answered by
            Ana
            
    i tried to insert .6447721757 and the site says round-off error.
    
                    Answered by
            Matt
            
    Damon, I think you made an arithmetic error here: 3.75 m g = 44.5 d^2 - .85 m g
44.5 d^2 = 2.90 m g
It should be 44.5d^2 = 4.6 m g
Then solve from there, I made that change and got the right answer.
    
44.5 d^2 = 2.90 m g
It should be 44.5d^2 = 4.6 m g
Then solve from there, I made that change and got the right answer.
                    Answered by
            Ana
            
    44.5 d^2=4.6mg
d^2=(4.6*.5*9.81)/(44.5)=0.507033708
d=0.712062994m
Is this right? I have only one chance to insert the answer so i want to make sure.
    
d^2=(4.6*.5*9.81)/(44.5)=0.507033708
d=0.712062994m
Is this right? I have only one chance to insert the answer so i want to make sure.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.