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?
What is her kinetic energy? What will happen to that kinetic energy and she swing upward?
#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?
Since there is no friction, all of the gravitational PE loss will go into the KE gain and the fact that the angle of the incline is 35° will not be relevant. If there was friction, then the angle of the incline would be relevant to the calculation of the frictional force and the work done against this dissipative force.
#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
You didn't square the displacement. That should take care of it. Note that the - sign, if applied, results in an imaginary velocity (square root of the negative)
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." ""
Good questions. Sorry for the delay in my reply.