Question

The force exerted by a stretched chord at given displacements is shown in the table given below. Experimentally, the force is found to vary proportionally to the square of the displacement, i.e. $F\left(x\right)=-h{x}^2$ where $h$ is some constant.
If the potential energy at $x=0\text{m}$ is $U_0=0\text{J}$, determine the potential energy at $x=1.5\text{m}$.

• A. $1.13\text{J}$
• B. $2.25\text{J}$
• C. $4.50\text{J}$
• D. $6.75\text{J}$

Answer 1

The correct option is B $2.25\text{J}$

The force exerted by chord $F=-h{x}^2$
Given: $F=-2\text{N}$ at $x=1\text{m}$
$\therefore -2=-h\times 1^2\Rightarrow h=2$
Thus the force is $F=-2x^2\text{N}$
Potential energy $U=\int -Fdx=\int 2x^{2}dx=\frac{2}{3}{x}^3+C$
Given $U=0$ at $x=0$ $\Rightarrow C=0$
$\therefore U=\frac{2}{3}{x}^3$
${\begin{array}{c}U\end{array}|}_{x=1.5}=\frac{2}{3}\times (1.5{)}^3=2.25\text{J}$