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Math Help - deceleration and forces

  1. #1
    Senior Member furor celtica's Avatar
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    deceleration and forces

    ok i just need an explanation for one tiny thing.

    a drop-forge hammer of mass 1500kg falls under gravity on to a piece of hot metal which rests in a fixed die. from the instant the hammer strikes the piece of metal until it comes to rest, the hammer is decelerating at 1.5 ms^2.
    find the magnitude of the force exerted by the hammer on the piece of metal while (a.) the hammer is decelerating (b.) after the hammer has come to rest.

    of course b. was easy, 15000N
    for a. i'm not sure, i know it has something to do with newton's third law. the answer is 17500 N but why would the force be greater while the hammer is decelerating? isn't deceleration just the difference between opposing forces divided by weight? in this case i would have answered that the force was 15000N as well.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    The force that the hammer exerts on the hot metal is given by:

    F = \dfrac{dp}{dt}

    Initial momentum is 1500v
    Final momentum is 0.

    This gives:

    F = \dfrac{(1500v-0)}{dt} = 1500 \dfrac{dv}{dt}

    Where dv/dt is acceleration, being negative here.

    So, net force = mg + F = 15 000 + 1 250 = 17 250 N

    Maybe there's a typo in the answer you gave? Or if it's to the nearest 500 N, then it becomes 17 500 N
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  3. #3
    Senior Member furor celtica's Avatar
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    typo, thanks
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  4. #4
    Senior Member
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    Quote Originally Posted by furor celtica View Post
    OK, i just need an explanation for one tiny thing.

    a drop-forge hammer of mass 1500kg falls under gravity on to a piece of hot metal which rests in a fixed die. from the instant the hammer strikes the piece of metal until it comes to rest, the hammer is decelerating at 1.5 m/s^2.
    find the magnitude of the force exerted by the hammer on the piece of metal while (a.) the hammer is decelerating (b.) after the hammer has come to rest.

    of course b. was easy, 15000N
    for a. I'm not sure, i know it has something to do with newton's third law. the answer is 17500 N but why would the force be greater while the hammer is decelerating? (because, not only does the metal support the hammer, it's applying the force to slow the hammer.) isn't deceleration just the difference between opposing forces divided by weight? in this case i would have answered that the force was 15000N as well.
    What is the net force on the hammer while if decelerates at 1.5 m/s^2 ? (Use Newton's 2nd Law.) \vec{F}_{net}=m\,\vec{a}. Note: The acceleration is upward.

    What two forces make up the net force on the hammer, while it decelerates?
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