## 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

1. anonymous

@Michele_Laino one more for you!

2. 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: $\Large L = - \Delta KE$

3. shamim

This is work energy theorem

4. 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

5. shamim

So work done w=change of kinetic energy

6. Michele_Laino

the work done by the forces which are acting on the particle