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.

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

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.

Beats me, but check my arithmetic carefully.

i tried to insert .6447721757 and the site says round-off error.

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=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.

It seems like you are on the right track with setting up the equation using the conservation of mechanical energy. However, there could be a mistake in your calculation.

Let's go through the steps to derive the correct equation:

1. The potential energy stored in the spring is given by: U_spring = (1/2)kx^2, where k is the spring constant and x is the displacement of the spring.

2. The initial potential energy stored in the spring is equal to the initial gravitational potential energy of the block: (1/2)kx^2 = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the initial height of the block.

3. The initial height of the block can be determined from the compression distance, d, and the radius of the loop, R. Using trigonometry, we have: h = R(1 - cos(theta)), where theta is the angle of the loop.

4. The work done by friction along the track AB is given by: W_friction = -µkmgd, where µk is the coefficient of kinetic friction and d is the length of AB.

5. The work done by the gravitational force along the loop is given by: W_gravity = mgh.

6. The total mechanical energy at the top of the loop is equal to the sum of the initial potential energy and the work done by gravity: (1/2)kx^2 + W_gravity = (1/2)mv^2 + (1/2)mv^2, where v is the velocity of the block at the top of the loop.

7. The velocity of the block at the top of the loop can be obtained using the conservation of mechanical energy. The gravitational potential energy is converted to kinetic energy: mgh = (1/2)mv^2.

8. Rearrange the equation to solve for v: v = sqrt(2gh).

9. Substitute the expression for v into the equation from step 6 and solve for x.

By following these steps, you should be able to determine the minimum compression distance, d, of the spring that enables the block to just make it through the loop-the-loop at point C.