N and 
N act together on a particle…
by Problem:
The constant forces 
and 
act together
on a particle during a displacement from position 
to position 
. Determine the total work done on the particle.
Solution:
Step 1: Calculate the total force
The total force 
is the sum of the two forces:
![\[\mathbf{F} = \mathbf{F_1} + \mathbf{F_2}\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-5da6011b2608a913079cda3ab7d46527_l3.png "Rendered by QuickLaTeX.com")
Calculation:
![\[ \mathbf{F} &= (\hat{i} + 2\hat{j} + 3\hat{k}) + (4\hat{i} - 5\hat{j} - 2\hat{k}) \]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-f81edea2ebf0d9d2ecb0bb65b70d7a1f_l3.png "Rendered by QuickLaTeX.com")
![\[= (\hat{i} + 4\hat{i}) + (2\hat{j} - 5\hat{j}) + (3\hat{k} - 2\hat{k}) \]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-35b0fc9a36ac08ad873ef096d8f4e2e2_l3.png "Rendered by QuickLaTeX.com")
![\[= 5\hat{i} - 3\hat{j} + \hat{k} \, \text{N} \]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-adaf3faa31b2a2f9242409d65f135bd5_l3.png "Rendered by QuickLaTeX.com")
Explanation: The components in the same directions (
) are added to get the total force.
Step 2: Calculate the Displacement Vector
The displacement vector 
is obtained by subtracting the initial position from the final position:
![\[\mathbf{s} = \mathbf{r_1} - \mathbf{r_2}\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-99962ef667191995c4f9e7da04c69aeb_l3.png "Rendered by QuickLaTeX.com")
initial position:
![\[\mathbf{r_2} = 0\hat{i} + 0\hat{j} + 7\hat{k} \, \text{cm}\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-276953912f6ab65c1eb9d6a5131d721d_l3.png "Rendered by QuickLaTeX.com")
Final position:
![\[\mathbf{r_1} = 20\hat{i} + 15\hat{j} + 0\hat{k} \, \text{cm}\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-b8532eb4c9368c27f8ac588c94e3bad5_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\[\mathbf{s} &= (20\hat{i} + 15\hat{j} + 0\hat{k}) - (0\hat{i} + 0\hat{j} + 7\hat{k})\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-29b488507e70688276eccf5beaff3447_l3.png "Rendered by QuickLaTeX.com")
![\[= (20 - 0)\hat{i} + (15 - 0)\hat{j} + (0 - 7)\hat{k}\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-8d77f7552b20df49df3f939f4caef711_l3.png "Rendered by QuickLaTeX.com")
![\[= 20\hat{i} + 15\hat{j} - 7\hat{k} \, \text{cm} \]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-a559020928c5cb14e455f5d13ab8f171_l3.png "Rendered by QuickLaTeX.com")
Convert this to meters (since the force is in Newtons):
![\[\mathbf{s} = 0.20\hat{i} + 0.15\hat{j} - 0.07\hat{k} \, \text{m}\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-ae57f5af556d89d65d367f337502c9fd_l3.png "Rendered by QuickLaTeX.com")
Explanation: The displacement vector shows the direction and distance in which the particle is displaced.
You may also want to know about projectile motion.
Step 3: Calculate the work
The formula for work 
is:
![\[W = \mathbf{F} \cdot \mathbf{s}\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-0c49fa9108255beb640b5ee9e20f5512_l3.png "Rendered by QuickLaTeX.com")
Here “
” represents the dot product.
Calculation:
![\[ W &= (5\hat{i} - 3\hat{j} + \hat{k}) \cdot (0.20\hat{i} + 0.15\hat{j} - 0.07\hat{k})\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-eaa799168179e79c6c2815f68f022e39_l3.png "Rendered by QuickLaTeX.com")
![\[= (5 \times 0.20) + (-3 \times 0.15) + (1 \times -0.07)\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-0eefd4877374da5e86551f9f2d90ad04_l3.png "Rendered by QuickLaTeX.com")
![\[= 1.00 - 0.45 - 0.07 \\&= 0.48 \, \text{Joule}\]](https://physicsqanda.com/wp-content/ql-cache/quicklatex.com-0a2eb2cbda333ce4ef082e7c6eb685ec_l3.png "Rendered by QuickLaTeX.com")
Explanation: In dot product, components in same directions are multiplied and then added. This ensures correct calculation of work as work is scalar product between force and displacement.
Final Answer:
The total work is 
joules.
References:
