VitaX

Most popular questions and responses by VitaX
  1. Genereal Physics I

    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 other end is fixed. The ladle has a kinetic energy of 260 J as it passes through its equilibrium position (the point at which the

    asked on October 13, 2010
  2. General Physics I

    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 work is: I found this question online so I saw the answer is 2500 Watts. What I don't understand is how to come to this

    asked on October 13, 2010
  3. physics

    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. The block sticks to the spring and momentarily stops after compressing the spring 18.4 cm. How much work is done (a) by the block

    asked on October 16, 2010
  4. physics

    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 horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the

    asked on October 16, 2010
  1. physics

    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

    posted on October 17, 2010
  2. physics

    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?

    posted on October 17, 2010
  3. physics

    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?

    posted on October 16, 2010
  4. physics

    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?

    posted on October 16, 2010
  5. Genereal Physics I

    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

    posted on October 13, 2010
  6. Genereal Physics I

    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

    posted on October 13, 2010
  7. General Physics I

    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

    posted on October 13, 2010