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Mid-term tomorrow, need some checking of answers of review packet

  1. Jan 25, 2005 #1
    Basically my teacher hands us out a review packet for physics and says to do it if we want practice...Okay I do the packet and I have it done, now my problem is that the teacher didn't give out the answers to the problems so I have no idea whether or not I did it right...so Im hoping some one will check my work before I take my midterm tomorrow. Sorry for being so troublesome.

    Anyways here are a few problems and my work involved. I don't know how to do those picture equations so bear with me

    --------------------------
    1. The height of a helicopter above ground is given by h = 3t³, where h is in meters and t is in seconds. After 2 seconds, the helicopter releases a small mailbag, How long after its release does the mailbag reach the ground.

    h = 3t³
    h(2) = 24 m

    h' = 9t²
    v(2) = h' = (9)(2)² = 36 m/s

    Si 0
    Sf 24m
    Vi -36 m/s
    Vf 42 m/s
    a 9.8 m/s²
    t ?

    I know that I have my signs placed oddly but thats how my brain works. Also I put all the information I have in the chart already from the work I get below except time.

    Vf² = Vi² + 2a(Sf - Si)
    Vf = sqrt(Vi² + 2a(Sf-Si))
    Vf = sqrt((-36)² + (2)(9.8)(24))
    Vf = 42 m/s

    Vf = Vi + at
    t = Vf-Vi/a
    t = 42+36/9.8

    t = 7.96 s

    I think thats right but I wanna make sure.
    -------------------------------
    2. A 100 g bullet is fired from a rifle having a barrel 0.6 m long. Assuming the origin is placed where the bullet begins to move, the force (in Newtons) exerted on the bullet by the expanding gas is 15000 + 10000x - 25000x², where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

    work:

    F(x) = 15000 + 10000x - 25000x²
    F(.6) = 15000 + 10000(.6) - 25000(.6)²
    F(.6) = 12000 N

    w = Fd
    w = (12000N)(.6N) = 7200 J

    Okay I think I did that problem horribly wrong because I mean its too easy and this is advanced physics.

    One last problem
    ----------------------------
    A catcher "gives" with the ball when he catches a 0.15 kg baseball moving at 25 m/s. If he moves his glove a distance of 2 cm, what is the average force acating on his hand?

    Work

    Ke = (1/2)mv²
    Ke = (1/2)(.15)(25)²
    Ke = 46.87 J

    Ke = (1/2)(.15)(0)² (Im assuming the ball stops)
    Ke = 0 J

    w = Kef - Kei
    w = Fd
    F = w/d
    F = 46.87J/.02m = 2343.75 N

    thanks for your time...I feel really bad now asking someone to check my work.
     
  2. jcsd
  3. Jan 25, 2005 #2
    1. Alright, looks right you did it right to begin with. The height the the velocity are correct. But I think you unnecessarily complicated the second half. Recall that
    sf - si = vi*t + (1/2)at^2

    In this case, a is going to be -9.8. Just use that and use the quadratic to solve for t directly. However, i do believe you answer is still right. So i guess, whatever works for you.

    2. You're right, it did seem too easy. dW/dx = F. Therefore, what is work in this situation?

    3. I think its ok.
     
  4. Jan 25, 2005 #3
    okay for number 2

    F(x) = 15000 + 10000x - 25000x²
    F'(x) = 10000 - 50000x
    F'(.6) = 10000 - 50000(.6)
    F'(6) = -20000

    w = fd
    w = -(20000)(.6)
    w = -12000 J

    err I don't think you can have negative Joules...
     
  5. Jan 25, 2005 #4
    wait

    so w = Integral(F)
    ?

    that seems a little more farfetched... since thats still a pretty big equation..
     
  6. Jan 25, 2005 #5
    As farfetched as it may seem, that is true. W = integral(F). But why did you say w = (-20000)(0.6). You actually already calculated the work when you solve your integral. You're right, typically you wouldnt have negative energy. But the negative sign really doesn't mean anything to you. For example, lets say a force acted on something for 2 meters. Than the work is 2F. Now if it was -2 m, than it would be -2F. The two statements are completely identical, in this situation also, the negative sign wouldn't mean much.
     
  7. Jan 25, 2005 #6
    Oh sorry, i made a mistake. You were supposed to take the integral, as you suggested in your second post. Than simply solve for the work at 0.6 m.
     
  8. Jan 25, 2005 #7
    so its

    F = 15000X + 5000X² - (25000/3)x³

    F(.6) = 15000(.6) + 5000(.6²) - (25000/3)(.6³)

    F = 9000N
    W = fd
    w = 9000(.6)
    w = 5400 J
     
  9. Jan 25, 2005 #8
    err okay a friend told me im not supposed to put it into w = fd because by solving for the intergral we get work since force is a factor or distance...is the person correct?
     
  10. Jan 26, 2005 #9
    Yea, your friend is correct.
     
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