Question

A $10kg$ block placed on a rough horizontal floor is being pulley by a constant force of $50N$. Coefficient of kinetic friction between the block and the floor is $0.4$. Find the work done by each individual force acting on the block over a displacement of $5m$.

Answer 1

Net force acting on the block:

$F-f=ma$

$50-\mu mg=ma$

$50-0.44\times 10\times 10=ma$

$50-40=ma$

$a=\frac{10}{10}=1m/s^2$

Net force $50-40=10N$

Work done by constant force:=F$\times$ displacement

$=50\times 5=250J$

Work done by friction:

$=-f\times 5$

$=-40\times 5=-200J$

It is negative since displacement is opposite to the direction of friction.

Net work$=250-200=50J$