Question #87918
Calculate the work done by a force F i j kˆ 2 ˆ ( ) ˆ 3 2 = x + xz − y + z r in moving a particle
along the curve x = t y = t z = t − t 2 2 2 from ; ; 4 t = 0 to t = 1. Is the force conservative?
1
Expert's answer
2019-04-12T09:05:07-0400

Force field


F=3x2i+(2xzy)j+zkF=3x^2\overrightarrow{i}+(2xz-y)\overrightarrow{j}+z\overrightarrow{k}

Work done along the space curve x=2t2,y=t,z=4t2t,from t=0 to t=1x=2t^2, y=t, z=4t^2-t, from \ t=0\ to\ t=1


Work=Fdr=01Fdrdtdt=Work=\int \overrightarrow{F}\cdot d\overrightarrow{r}=\displaystyle\int_{0}^1\overrightarrow{F}\cdot {d\overrightarrow{r} \over dt} dt=

=01[3(2t2)2i+(2(2t2)(4t2t)t)j+(4t2t)k][4ti+1j+(8t1)k]dt==\displaystyle\int_{0}^1[3(2t^2)^2\overrightarrow{i}+(2(2t^2)(4t^2-t)-t)\overrightarrow{j}+(4t^2-t)\overrightarrow{k}]\cdot [4t\overrightarrow{i}+1\overrightarrow{j}+(8t-1)\overrightarrow{k}]dt=

=01[48t5+16t44t3t+32t34t28t2+t]dr==\displaystyle\int_{0}^1[48t^5+16t^4-4t^3-t+32t^3-4t^2-8t^2+t]dr=

=[8t6+165t5+7t44t3]10=8+165+740=715=[8t^6+{16 \over 5}t^5+7t^4-4t^3]\begin{matrix} 1 \\ 0 \end{matrix}=8+{16 \over 5}+7-4-0={71 \over 5}

Work done along the straight line x=2t,y=t,z=3t,from t=0 to t=1x=2t, y=t, z=3t, from \ t=0\ to\ t=1


Work=Fdr=01Fdrdtdt=Work=\int \overrightarrow{F}\cdot d\overrightarrow{r}=\displaystyle\int_{0}^1\overrightarrow{F}\cdot {d\overrightarrow{r} \over dt} dt=

=01[3(2t)2i+(2(2t)(3t)t)j+3tk][2i+1j+3k]dt==\displaystyle\int_{0}^1[3(2t)^2\overrightarrow{i}+(2(2t)(3t)-t)\overrightarrow{j}+3t\overrightarrow{k}]\cdot [2\overrightarrow{i}+1\overrightarrow{j}+3\overrightarrow{k}]dt=

=01[24t2+12t2t+9t]dr==\displaystyle\int_{0}^1[24t^2+12t^2-t+9t]dr=

=[12t3+4t2]10=12+40=16≠715=[12t^3+4t^2]\begin{matrix} 1 \\ 0 \end{matrix}=12+4-0=16=\not {71 \over 5}

A conservative force is a force with the property that the total work done in moving a particle between two points is independent of the taken path.

Therefore, the force is not conservative.


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