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Questions (4)
A 506 g block is released from rest at height h0 above a vertical spring with spring constant k = 500 N/m and negligible mass.
4 answers
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A 0.720 kg snowball is fired from a cliff 8.10 m high. The snowball's initial velocity is 14.2 m/s, directed 30.0° above the
2 answers
899 views
A 0.49 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 640 N/m) whose
2 answers
1,288 views
A 50-N force acts on a 2-kg crate that starts from rest. When the force has been acting for 2 s the rate at which it is doing
3 answers
8,150 views
Answers (7)
I just tried 8.464 = mg(h-.184) and my answer for the height was still wrong. What am I doing wrong here ... I should add for part d) I have it set up so that mgh = .5kx^2 to find the max compression at 5h. I think both of these are right yet my answers
I think I screwed up somewhere because my values aren't correct. Am I supposed to set it up like mg(h-.184) = 8.464?
So tbe work done by the spring is the Potential Energy = mgh? I get 1.7069 m. For part d) How would you find the compression of the spring at h0 = 8.5345 m?
How can that be because the height at the ground is 0. So mgyf would be zero as well wouldn't it? How is my work for part a and b?
Ok, I figured it out. I find the potential energy, and subtract the potential energy from the total energy which was given in the problem. With that kinetic energy now I solve for V using kinetic energy equation. From there I'm able to find the Power with
I worked on this somewhat and I came to the conclusion that in part a since it's asking for the Power as it passes through its equilibrium position, the force is 0. Which makes the Power 0 Watts. For part b though I'm a little more confused as to how to
I was able to figure some things out. Like a = F/m = 50/2 = 25 m/s^2 s=Vo +.5at^2 = .5*25*2^2 = 50 m Work = 50*50 = 2500 J Power = Work/time = 2500/2 = 1250 W Problem is that the answer is 2500, but the multiple choice answers are all in Watts so that has
