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anonymous

  • one year ago

A 100 kg car is moving at a speed of 10 meter per second and come to rest after covering a distance of 50m. The amount of work done against friction is: (a) +5*10^1 J (b) +5*10^2 J (c) +5*10^3 J (d) +5*10^4 J

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  1. anonymous
    • one year ago
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    @Michele_Laino one more for you!

  2. Michele_Laino
    • one year ago
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    the kinetic energy change is: \[\Large \begin{gathered} \Delta KE = 0 - \frac{1}{2}m{v^2} = \hfill \\ \hfill \\ = - \frac{1}{2} \times 100 \times 100 = - 5 \times {10^3}Joules \hfill \\ \end{gathered} \] and the requested work, is such that: \[\Large L = - \Delta KE\]

  3. shamim
    • one year ago
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    This is work energy theorem

  4. Michele_Laino
    • one year ago
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    yes! More precisely, the work W done by friction forces is: \[\Large W = \Delta KE\] and that formula is the expression of the work energy theorem

  5. shamim
    • one year ago
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    So work done w=change of kinetic energy

  6. Michele_Laino
    • one year ago
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    the work done by the forces which are acting on the particle

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