[SRMcDonald]

Betty weighs 420 N and she is sitting on a playground swing seat that hangs 0.40 m above the ground. Tom pulls the swing back and releases it when the seat is 1.00 m above the ground.

a. How fast is Betty moving when the swing passes through its lowest position?

b. If Betty moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction?

a. 0 + mgh = 1/2mv^2 + 0
1/2mv^2 = mgh
v=sqrt(2gh) = 3.43 m/s

b. I'm not sure how to do this, so I had to hazard a guess:

KE top + PE top = KE bottom + PE bottom + Work of friction

=> 0 + mgh = 1/2mv^2 + 0 + Wf
=> mgh - 1/2mv^2 = Wf
=> (420 N)(0.60 m) - 1/2(420 N/9.80 m/s^2)(2.0 m/s)^2 = Wf

= 166 J

Where is the friction coming from? Is it air resistance against the swing that's accelerating downwards? Then wouldn't the work be negative since it's working upwards while she's still traveling downwards? I'm kind of confused here; I'd really appreciate the help.

I did my answers sloppily so if you don't know what I'm doing, I can edit it.

Thanks in advance. :)

[ebaines]

You have correct answers for both parts - good work! The fact that the swing is only going at 2 m/s, whereas in an ideal case she would be moving at 3.43 m/s, tells you that she has lost kinetic energy somehow. Something did work to slow her down - it could be air resistance as you surmise. However, keep in mind that work is always positive - it's not a vector, and there is no such thing as negative work. So always treat work and KE as positive numbers (there may be a change in kinetic enery that's negative, but the KE of an object is always positive).