Question

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

Answer 1

@Michele_Laino one more for you!

Answer 2

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:
\[\Large L = - \Delta KE\]

Answer 3

This is work energy theorem

Answer 4

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

Answer 5

So work done w=change of kinetic energy

Answer 6

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