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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino one more for you!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.1This is work energy theorem

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! 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
 one year ago
Best ResponseYou've already chosen the best response.1So work done w=change of kinetic energy

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the work done by the forces which are acting on the particle
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