Question

A particle is subjected to a force $F_x$ that varies with position as shown in figure. What is the total work done by the force over the distance $x=0$ to $x=15.0$ m?

• A. $-15J$
  
• B. $15J$
  
• C. $3J$
  
• D. $27J$
  

Answer 1

The correct option is D $27J$

The work done by the force is
$W={\int }_0^{15}Fdx={\int }_0^{4}Fdx+{\int }_4^{10}Fdx+{\int }_{10}^{12}Fdx+{\int }_{12}^{15}Fdx$
From the graph force
for $x=0$ to $x=4$, $\frac{F-0}{x-0}=\frac{0-3}{0-4}\Rightarrow F=\frac{3}{4}x$

for $x=4$ to $x=10,F=3$

for $x=10$ to $x=12$, $\frac{F-3}{x-10}=\frac{3-0}{10-12}\Rightarrow F=-\frac{3}{2}(x-10)+3$

for $x=12$ to $x=15,F=0$
so now, $W={\int }_0^4\frac{3}{4}xdx+{\int }_4^{10}3dx+{\int }_{10}^{12}[-\frac{3}{2}(x-10)+3]dx+0$
$=\frac{3}{4}\left(\frac{16}{2}\right)+3(10-4)-\frac{3}{2}[\frac{144-100}{2}-10(12-10\left)\right]+3(12-10)$
$=6+18-\frac{3}{2}(22-20)+6=24-3+6=27J$

Ans:(D)