Question

A $10\text{kg}$ block placed on a rough horizontal floor is being pulled by a constant force $F=100\text{N}$ acting at angle ${37}^{\circ }$. The coefficient of kinetic friction between the block and the floor is $0.4$. Find the total work done due to force $F$ acting on the block over the displacement of $5\text{m}$.

• A. $480\text{J}$
• B. $360\text{J}$
• C. $320\text{J}$
• D. $400\text{J}$

Answer 1

The correct option is C $320\text{J}$


Forces acting on block,
Weight or gravity force$\left(mg\right)=10\times 10\text{N}$$=100\text{N}$

Applied force along $x-$ direction,
$F_x=100\cos {\theta }^{\circ }=100\cos {37}^{\circ }=80\text{N}$


Applied force along $y-$ direction,
$F_y=100\sin {\theta }^{\circ }=100\sin {37}^{\circ }=60\text{N}$

In $y$ direction,
$N+F_y=mg=100\text{N}$
$N=100-60=40\text{N}$

$\therefore \text{Force of kinetic friction}\left(f\right)={\mu }_{k}N$
$=0.4\times 40\text{N}=16\text{N}$


Displacement$\Delta x=5$ m is along $x-$axis
$\therefore$ all the forces ${\perp }^r$ to displacement will have zero work.

$\therefore$ Total Work$\left(W_{\text{total}}\right)$ $=(F_x-f)\cdot \Delta x$
$=(80\text{N}-16\text{N})\cdot 5\text{m}$
$=320\text{J}$