course phy 121
Have I got questions for you!#33. Jane, looking for Tarzan, is running at top speed (5.3m/s) and grabs a vine hanging vertically froma tall tree the jungle. How high can she swing upward?
Can you just give me a clue as to where to start?
#34 A novice skier, starting from rest, slides down a frictionless 35 degree incline whose vertical height is 185m. How fast is she going when she reaches the bottom?
I used 1/2mv0^2 + mgy1 = 1/2mvf^2 + mgy2 and came up with the fairly reasonable answer of 60.22m/s. I don't know if it is correct since it is even numbered, but my concern is that I didn't use the piece of information about it being a 35degree incline. Did I approach this one wrong?
#35- I actually got this one right.
#39 A vertical spring (ignore mass), whose spring stiffness constant is 950N/m, is attached to a table and is compressed down 0.150m. What upward speed can it give to a .30-kg ball when released? How high above it's original position (spring compressed) will the ball fly?
I tried KE1 + PE1 = KE2 + PE2 and chose up as positive
1/2mv0^2 + 1/2kx1^2 = 1/2mvf^2 + 1/2kx2^2
0 + 1/2(950N/m)(-0.150m)= 1/2 (.30kg)(vf^2) + 0
but my vf comes out to be 21.79m/s and the book says it is 8.3m/s
So, then I thought I didn't account for gravity's effect on the ball, so I added mgy1 and mgy2 to the either side of the equation and I still came up with 21.something m/s as my answer. I have tried changing positive and negative all over the place and I think I am plugging in the right numbers into the equation, so am I using the wrong equation? What have I done wrong?
Then to figure out the distance should I use vf^2 = v0^2 + 2a*ds? Since I can't even get the right velocity, I'm not too worried about the distance yet.
Thanks for any help you can give." ""