Question

Kyle pushes a 50 kg sack of rice across a level floor by horizontal force of 35.0 N against a frictional force of 12.0 N. He succeeded in moving the sack to a distance of 5.0 m. How much work is done by (a) Kyle, (b) frictional, and (c) force of gravity?

Answer 1

Mass of sack=50 kg

Distance traveled by sack(s)=5 m

(a) Force applied by Kyle=$35N$

Work done by Kyle=F.s=35$\times5=175 J$

(b) Force by friction=$-12N$ (as frictional force work in opposite direction)

Work done by frictional force=$-12\times5=-60J$

(c) Work done by gravity=$F.s=Fs cos\theta=Fscos 90\degree=0N$

(As gravitational force act downward and displacement is in horizontal direction that is mutually perpendicular)