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Question: A particle is acted on by constant forces \({\text{ = - 3i + 2j + 5k}}\) and \({\text{2i + j + - 3k}...

A particle is acted on by constant forces  = - 3i + 2j + 5k{\text{ = - 3i + 2j + 5k}} and 2i + j + - 3k{\text{2i + j + - 3k}} and is displaced from point (2,1,3)\left( {2, - 1, - 3} \right) to the point (4,3,7)\left( {4, - 3,7} \right). Find the total work done by the forces.

Explanation

Solution

To solve this we have to apply a work done formula which is the product of force and displacement. Here as two forces are given to us we first calculate the total forces acting on the particle then find the final displacement vector. Now find the product of total force and final displacement we will get our solution.

Complete step by step answer:
As from the problem we have a particle that acted on by constant forces  = - 3i + 2j + 5k{\text{ = - 3i + 2j + 5k}} and 2i + j + - 3k{\text{2i + j + - 3k}} and is displaced from point (2,1,3)\left( {2, - 1, - 3} \right) to the point (4,3,7)\left( {4, - 3,7} \right).
Now from the above given values we have to find the total work done by the forces.
We know the total work done by a force F\overrightarrow F through a displacement d\overrightarrow d is given by,
W=FdW = \overrightarrow F \cdot \overrightarrow d

Now we know that,
F1= - 3i + 2j + 5k\overrightarrow {{F_1}} = {\text{ - 3i + 2j + 5k}}
And F2=2i + j + - 3k\overrightarrow {{F_2}} = {\text{2i + j + - 3k}}
Hence the total force F\overrightarrow F will be,
F=F1+F2\overrightarrow F = \overrightarrow {{F_1}} + \overrightarrow {{F_2}}
Now putting the respective value we will get,
F=( - 3i + 2j + 5k)+(2i + j + - 3k)\overrightarrow F = \left( {{\text{ - 3i + 2j + 5k}}} \right) + \left( {{\text{2i + j + - 3k}}} \right)
F= - i + 3j + 2k\Rightarrow \overrightarrow F = {\text{ - i + 3j + 2k}}
Now we know that,
d1=(2,1,3)=2i - 1j - 3k\overrightarrow {{d_1}} = \left( {2, - 1, - 3} \right) = {\text{2i - 1j - 3k}}
And d2=(4,3,7)=4i - 3j + 7k\overrightarrow {{d_2}} = \left( {4, - 3,7} \right) = {\text{4i - 3j + 7k}}

Hence the final displacement d\overrightarrow d will be,
d=d2d1\overrightarrow d = \overrightarrow {{d_2}} - \overrightarrow {{d_1}}
Now on putting the respective values we will get,
d=(4i - 3j + 7k)(2i - 1j - 3k)\overrightarrow d = \left( {{\text{4i - 3j + 7k}}} \right) - \left( {{\text{2i - 1j - 3k}}} \right)
d=2i - 2j + 10k\Rightarrow \overrightarrow d = {\text{2i - 2j + 10k}}
Now using work done formula we will get,
W=( - i + 3j + 2k)(2i - 2j + 10k)W = \left( {{\text{ - i + 3j + 2k}}} \right) \cdot \left( {{\text{2i - 2j + 10k}}} \right)
Now using dot product we will get,
W=(1×2)+(3×2)+(2×10)=12J\therefore W = \left( { - 1 \times 2} \right) + \left( {3 \times - 2} \right) + \left( {2 \times 10} \right)=12\,J

Hence the work done by the forces is W=12JW = 12J.

Note: The SI unit of work is joules. Work is said to be done whenever a force moves something over a distance with the application of two or more forces. Here we have used the dot product of vectors to find the work done which is the sum of the product of the corresponding entries of the two sequences of numbers.