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anonymous

  • 4 years ago

A 1.2 x 10^4 kg truck is travelling south at 22 m/s. (a) what net force is required to bring the truck to a stop in 330m?

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  1. JamesJ
    • 4 years ago
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    By the Work-Energy theorem, the amount of work done on the truck is equal to its change in energy. Work, W, is given by W = Fd where F is a force applied in the direction of a distance d. Now, this must be equal to the change in KE. See what to do now?

  2. JamesJ
    • 4 years ago
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    What is the change in KE for the truck?

  3. anonymous
    • 4 years ago
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    is it possible to use \[v_{f}^{2} = v_{i}^{2} + 2ad \] to get the answer to this question? because that's what my friend told me to use. and then she told me to use\[a= \left( v _{f}^{2} - v_{i}^{2}\right) / 2d\] but then i want to know how she got from the first equation to the second equation...

  4. JamesJ
    • 4 years ago
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    It is ... let me come back to it. Follow my logic for now. What is the expression for KE?

  5. JamesJ
    • 4 years ago
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    \[ KE = \frac{1}{2}mv^2 \] where v is speed

  6. JamesJ
    • 4 years ago
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    Hence change in KE is \[ \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \] By the Work-Energy theorem, this change in KE must be equal to the amount of work done on the truck. That work is given by \[ Work = Fd = mad \] because F = ma

  7. JamesJ
    • 4 years ago
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    Therefore, as we have by the Work-Energy Theorem: \[ Work = \Delta KE \] this means \[ Fd = \frac{1}{2}m (v_f^2 - v_i^2) \] You know that v_f = 0, therefore \[ Fd = - \frac{1}{2} m v_i^2 \] You know d, m, and v_i. Hence you can solve for F.

  8. JamesJ
    • 4 years ago
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    What we've done here is how we derive the other equation your friend is referring to. By Newton's second law, F = ma, hence Work = Fd = mad thus \[ mad = \frac{1}{2} m (v_f^2 - v_i^2) \] Cancel m from both sides, multiply through by 2 and we have \[ 2ad = v_f^2 - v_i^2 \] From that, you could also solve for acceleration a, and therefore calculate F using F = ma. But the first derivation I've written above is more fundamental, and better reflects the Physics of the situation.

  9. anonymous
    • 4 years ago
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    Oh, that makes more sense. i got \[a= -0.733\] and when i substitute that to \[F=ma\] i get \[F=(1.2x10^{2} kg)(-0.733)\] \[F=-8800N\] F=8800 N [N] \[F= 8.8 \times 10^{3} \] N [N]

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