Question

Find the total work done in moving a particle in a force field given by 𝐹 = 3𝑥𝑦𝑖 − 5𝑧𝑗 + 10𝑥𝑘 along the curve x = 𝑡 2 + 1, 𝑦 = 2𝑡2 , 𝑧 = 𝑡 3 𝑓𝑟𝑜𝑚 𝑡 = 1 𝑡𝑜 𝑡 = 2.

Answer 1

The total work done can be found as the integral of the scalar product between the force vector F and a certain displacement trajectory vector ds. We use the definition for the curve r as well to proceed:

$\vec{r} = \begin{pmatrix} t^2+1, &2t^2, & t^3 \end{pmatrix}$ (between t=1 and t=2) which gives$\frac{d\vec{r}}{dt} = \begin{pmatrix} 2t, &4t, & 3t^2 \end{pmatrix}$

$\vec{F} = \begin{pmatrix} 3xy, &-5z, & 10x \end{pmatrix}$ which we have to substitute in terms of the 't' variable:

$\vec{F} =\begin{pmatrix} 3(t^2+1)(2t^2), &-5(t^3), & 10(t^2+1) \end{pmatrix}$

$\vec{F} = \begin{pmatrix} 6t^4+6t^2, &-5t^3, & 10+10t^2 \end{pmatrix}$

Then we substitute and calculate the total work:

$W=\int_{C}^{} \vec{F}\cdot d\vec{s} =\int_{C}^{} \vec{F}(\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}) dt$

$W = \int_{1}^{2} \begin{pmatrix} 6t^4+6t^2, &-5t^3, & 10+10t^2 \end{pmatrix} \cdot \begin{pmatrix} 2t, &4t, & 3t^2 \end{pmatrix} dt$

$=\int_{1}^{2} (12t^{5}+12t^{3}-20t^{4}+30t^{2}+30t^{4})dt$

$W=\int_{1}^{2} (12t^{5}+10t^{4}+12t^{3}+30t^{2})dt=\left [ 2t^{6}+2t^{5} +3t^{4}+10t^{3}\right ]_{1}^{2}= (2(2^6)+2(2^5)+3(2^4)+10(2^3))-(2(1^6)+2(1^5)+3(1^4)+10(1^3))=(128+64+48+80)-(2+2)+3+10)$

$W = (128+64+48+80)-(2+2+3+10) =320-17$

$W = 303\,u$

In conclusion, the total work done is 303 units.

Reference:

• Young, H. D., Freedman, R. A., & Ford, A. L. (2006). Sears and Zemansky's university physics (Vol. 1). Pearson education.