Question

suppose that you package with force 30 kg package 120 N through distance of 0.8 m and that the opposing friction force average is 5 N
Calculate a) work done by force b) work done by friction c) net work done
If the force was applied for 200 ms determine the final velocity that package will have after the removal of the force

Answer 1

Explanation:

Work done = F.d (dot product)

If work done and displacement is in opposite direction work is negative

a) force is parallel to displacement, W = Fxd = 120x0.8

b) force is antiparallel, W= -5x0.8

c) Add the two above

d)Net work done (c) = change in kinetic energy =1/2xmxv^2

Solve!