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Homework Help: Various Exam Review Questions

  1. Jan 26, 2004 #1
    I'm working on a small worksheet for exam review that I could get some extra credit on. I've finished the problems posted below, but would like to know if they've been done correctly. If you can help on either, please do.


    "[3] A 6.0kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant, horizontal force of 12N. (a) Find the work done on the block after it has moved 3.0m. (b) Find the final kinetic energy of the block. (c) Find the acceleration of the block. (d) Find the speed of the block."

    For part (a) I used W=F*d (W=12*3) to get W to equal W=36J. I then used F=m*a to find part (c), which I found to be 2m/s^2. I used the answer I found in part (c) to find part (d) by using v^2=2ad. I found (d) to equal v=3.5m/s. I then used the value I found in part (d) to answer part (b) by plugging in the velocity to KE=(1/2)mv^2, which I found KE to equal KE=36.75J. These answers don't really check out, however, and I'm not sure where I went wrong.


    "[5] In a particular crash test, an automobile of mass 1500kg collides with a wall. The initial and final velocities of the automobile are -15m/s and 2.6m/s respectively. If the collision lasts for 0.150s, find the impulse due to the collision and the average force exerted on the automobile."

    For the first part of the question (impulse) I used F(delta)t=m(Vf-Vi), which I calculated to be F(delta)t=264000kgm/s.

    For the second part (average force) I used the fact that F(delta)t=(delta)p where (delta)p is change in momentum. Changing the equation to F = (delta)p / (delta)t, I found F to equal (26400kgm/s)/(0.150s), or F=1.76x10^5N. Is this correct?

    Thanks for your time and help.
  2. jcsd
  3. Jan 26, 2004 #2

    Doc Al

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    Staff: Mentor

    You methods are fine, but you did a little more work than needed. The work done on the block equals the final KE, so parts a and b have the same answer. Your answers don't check out perfectly due to round off error: for example, I got v = 3.46, not 3.5.
    Yes. (You did have a typo in your answer to the first part.)
  4. Jan 26, 2004 #3
    Yep, looks like I added an extra zero on number 5 part one for good measure :wink:

    Thanks for the help here Doc, I have one more problem on the sheet that I need to look at that deals with circuits. Depending on how I do on that, it may end up here tomorrow. Thanks again for the help, sir.
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