# A worker pushes a wheelbarrow with a force 40N over a level distance 6m. Frictional force 24N acts on the wheelbarrow in opposite direction to that of the worker, find the net work done on wheelbarrow?

Aug 6, 2015

I found: $W = 96 J$
${F}_{\text{net}} = \Delta F = {F}_{1} - f = 40 - 24 = 16 N$
$W = {F}_{\text{net}} \cdot s = 16 \cdot 6 = 96 J$