anonymous
  • anonymous
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
Physics
katieb
  • katieb
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anonymous
  • anonymous
@Michele_Laino one more for you!
Michele_Laino
  • Michele_Laino
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: \[\Large L = - \Delta KE\]
shamim
  • shamim
This is work energy theorem

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Michele_Laino
  • Michele_Laino
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
shamim
  • shamim
So work done w=change of kinetic energy
Michele_Laino
  • Michele_Laino
the work done by the forces which are acting on the particle

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