For SBP #5, I get t = 2 * sqrt (L/3g).  The answer key has t = 2 * sqrt (L/g).  Have I missed something?

Genevieve, Use the second kinematics equation in the y direction:  y = y(zero) +v(zero y) -1/2 gt^2.  y = 0 y(zero) = 2L.  so:  2L =1/2 gt^2.   then t =2 *sqrt(L/g).  The x velocity is not used to find the time. John

For SBP #7 part A, I get 3.56 J, not 3.63J like the key says.  Here's what I did: total energy = gravitational PE at the top + kinetic energy at the top = mgh + 0.5mv^2 = (.1)(9.8)(1.8) + 0.5(.1)(6)^2 = 3.564
Am I overlooking something?

Savanna, I get the same answers as you.  I will correct the answer key.  Thanks, John

Also for SBP #7 part C)ii), I got 7.4N not 9.4N.  Here's what I did:  At the bottom, T - mg = mv^2/r, so T = mv^2/r -mg = m(v^2/r - g).  Then T = .1(8.2^2/0.8 - 9.8) = 7.4 N.

You are also right on this one.

Just looked again, and I was actually wrong.  I subtracted mg from mv^2/r, when it should have been addition.  I tried again and did get 9.4 N.4

At the bottom, T - mg = mv^2/r, so T = mv^2/r +mg = m(v^2/r + g).  Then T = .1(8.2^2/0.8 + 9.8) = 9.4 N.

correct - I made the same error. I fixed the answer key. The answer is 9.4 N.

I am confused why #7 is in this chapter at all. Does it not come before energy?

Christopher, you're correct. The focus of the problem is circular motion, and the assumption was that students might have had energy in an Algebra Based Physics course, and this was their second course. However, we are finding that this is not always the case, so this problem does belong in the energy unit. We'll leave it here for now, and teachers can come back to this problem after the energy unit, or can assign it if their students did have a previous course.  John