A massless spring of constant k =72.0 N/m is fixed on the left side of a level t

Question

A massless spring of constant k =72.0 N/m is fixed on the left side of a level track. A block of mass m = 0.4 kg is pressed against the spring and compresses it a distance d, as in the figure below. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.7 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 ?k=0.2, and that the length of AB is 1.9 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 exerted by the track on the block will be zero if the block barely makes it through the loop-the-loop.)

A massless spring of constant k =72.0 N/m is fixed on the left side of a level track. A block of mass m = 0.4 kg is pressed against the spring and compresses it a distance d, as in the figure below. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.7 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 ?k = 0.2, and that the length of AB is 1.9 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 exerted by the track on the block will be zero if the block barely makes it through the loop-the-loop.)

Explanation / Answer

This is a conservation of energy problem.

Remember that the total energy of a system remains constant. That is:

Ef = Ei (Final Energy = Initial Energy).

Initially, you have the potential energy stored in the spring.

Afterwards, you have the kinetic energy of the block, as well as gravitational potential energy of the block at the top of the loop-the-loop.

With regards to the portion of "Just making it through the loop-the-loop" is sort of vague. I take it to mean that the block should not lose contact with the loop-the-loop.

Recall that the equation for potential energy of a spring is:

1/2kd^2 (where k is the spring constant and d is the displacement) [normally x is used instead of d, but since you mentioned d above, I'll use d]

The equation for linear kinetic energy is:

1/2mv^2

And the equation for gravitational potential energy is:

mgh (where h is the height above zero*) *zero meaning the chosen point of zero potential energy. In this problem, we take the 'zero' position to be the height at which the block starts.

Finally, we have the energy lost due to friction. This is represented by the work done by friction:

Wf = ukNx (uk is the coefficient of kinetic friction, N is the normal force, and x is the displacement)

So setting up our equation for conservation of energy we have:

1/2kd^2 - ukmgx = 1/2mv^2 + mgh (recall N = mg)

k = 72 N/m
uk = 0.2
x =1.9m
m = 0.40kg
g = 9.8m/s^2
h = 3.4m (2 * R) - I take h to be the position at the top of the loop-the-loop, which means the diameter of the circle = 2R

So we just need to solve for D.

We are still missing velocity, so we need to deal with that as well.

To determine velocity, we need to consider the forces acting upon the block at point C.

There are two. N (normal force) and mg (weight). There is a third force generated by the block moving, but it is generated by the velocity. This is the centripetal force, and it is equal to mv^2/R.

The provision that it JUST makes it around the loop-the-loop implies that N = 0. Drawing a free-body-diagram allows you to set the centripetal force equation:

mg + N = mv^2/R (it is mg + N because they are both in the same direction.)

Since N = 0, we have:

mg = mv^2/r.

Simplify this and you get:

g = v^2/r

or

v = sqrt(rg).

Substitute this back into your first equation:

1/2kd^2 - ukmgx = 1/2m(rg) + mgh (recall v = sqrt(rg) so v^2 = rg)

You now only have one unknown, so solve for D.

Hope that helps!

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