Force field
F = 3 x 2 i → + ( 2 x z − y ) j → + z k → F=3x^2\overrightarrow{i}+(2xz-y)\overrightarrow{j}+z\overrightarrow{k} F = 3 x 2 i + ( 2 x z − y ) j + z k Work done along the space curve x = 2 t 2 , y = t , z = 4 t 2 − t , f r o m t = 0 t o t = 1 x=2t^2, y=t, z=4t^2-t, from \ t=0\ to\ t=1 x = 2 t 2 , y = t , z = 4 t 2 − t , f ro m t = 0 t o t = 1
W o r k = ∫ F → ⋅ d r → = ∫ 0 1 F → ⋅ d r → d t d t = Work=\int \overrightarrow{F}\cdot d\overrightarrow{r}=\displaystyle\int_{0}^1\overrightarrow{F}\cdot {d\overrightarrow{r} \over dt} dt= W or k = ∫ F ⋅ d r = ∫ 0 1 F ⋅ d t d r d t =
= ∫ 0 1 [ 3 ( 2 t 2 ) 2 i → + ( 2 ( 2 t 2 ) ( 4 t 2 − t ) − t ) j → + ( 4 t 2 − t ) k → ] ⋅ [ 4 t i → + 1 j → + ( 8 t − 1 ) k → ] d t = =\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= = ∫ 0 1 [ 3 ( 2 t 2 ) 2 i + ( 2 ( 2 t 2 ) ( 4 t 2 − t ) − t ) j + ( 4 t 2 − t ) k ] ⋅ [ 4 t i + 1 j + ( 8 t − 1 ) k ] d t =
= ∫ 0 1 [ 48 t 5 + 16 t 4 − 4 t 3 − t + 32 t 3 − 4 t 2 − 8 t 2 + t ] d r = =\displaystyle\int_{0}^1[48t^5+16t^4-4t^3-t+32t^3-4t^2-8t^2+t]dr= = ∫ 0 1 [ 48 t 5 + 16 t 4 − 4 t 3 − t + 32 t 3 − 4 t 2 − 8 t 2 + t ] d r =
= [ 8 t 6 + 16 5 t 5 + 7 t 4 − 4 t 3 ] 1 0 = 8 + 16 5 + 7 − 4 − 0 = 71 5 =[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} = [ 8 t 6 + 5 16 t 5 + 7 t 4 − 4 t 3 ] 1 0 = 8 + 5 16 + 7 − 4 − 0 = 5 71 Work done along the straight line x = 2 t , y = t , z = 3 t , f r o m t = 0 t o t = 1 x=2t, y=t, z=3t, from \ t=0\ to\ t=1 x = 2 t , y = t , z = 3 t , f ro m t = 0 t o t = 1
W o r k = ∫ F → ⋅ d r → = ∫ 0 1 F → ⋅ d r → d t d t = Work=\int \overrightarrow{F}\cdot d\overrightarrow{r}=\displaystyle\int_{0}^1\overrightarrow{F}\cdot {d\overrightarrow{r} \over dt} dt= W or k = ∫ F ⋅ d r = ∫ 0 1 F ⋅ d t d r d t =
= ∫ 0 1 [ 3 ( 2 t ) 2 i → + ( 2 ( 2 t ) ( 3 t ) − t ) j → + 3 t k → ] ⋅ [ 2 i → + 1 j → + 3 k → ] d t = =\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= = ∫ 0 1 [ 3 ( 2 t ) 2 i + ( 2 ( 2 t ) ( 3 t ) − t ) j + 3 t k ] ⋅ [ 2 i + 1 j + 3 k ] d t =
= ∫ 0 1 [ 24 t 2 + 12 t 2 − t + 9 t ] d r = =\displaystyle\int_{0}^1[24t^2+12t^2-t+9t]dr= = ∫ 0 1 [ 24 t 2 + 12 t 2 − t + 9 t ] d r =
= [ 12 t 3 + 4 t 2 ] 1 0 = 12 + 4 − 0 = 16 = ̸ 71 5 =[12t^3+4t^2]\begin{matrix}
1 \\
0
\end{matrix}=12+4-0=16=\not {71 \over 5} = [ 12 t 3 + 4 t 2 ] 1 0 = 12 + 4 − 0 = 16 = 5 71 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.
Comments