Question

Calculate the work done by the force $\overrightarrow{F}$ when the block moves from $x=0m$ to $x=6.0m$

• A. 4 J
• B. 0 J
• C. 2 J
• D. 3 J

Answer 1

The correct option is D 3 J

The work done by the force is $W={\int }_0^{7}Fdx={\int }_0^{2}Fdx+{\int }_2^{3}Fdx+{\int }_3^{4}Fdx+{\int }_4^{6}Fdx$
From the graph force
for $x=0$ to $x=2$, $\frac{F-0}{x-0}=\frac{0-2}{0-2}\Rightarrow F=x$
for $x=2$ to $x=3$, $F=2$
for $x=3$ to $x=4$, $F=0$
for $x=4$ to $x=6$, $\frac{F-0}{x-4}=\frac{0+1}{4-6}\Rightarrow F=-\frac{x}{2}+2$
now
$W={\int }_0^{2}xdx+{\int }_2^{3}2dx+0+{\int }_4^6(-\frac{x}{2}+2)dx+0$
$W=\left(\frac{4}{2}\right)+2(3-2)-\frac{1}{2}\left(\frac{36-16}{2}\right)+2(6-4)=2+2-5+4=3J$
Ans:(D)