Constant forces \(\rm \vec P\) = 2î - 5ĵ + 6k̂ and \(\rm \vec Q\) = -î + 2ĵ - k̂ act on a particle. The work done when the particle is displaced from A whose position vector is 4î - 3ĵ - 2k̂, to B whose position vector is 6î + ĵ - 3k̂, is:
-15 units.
Work Done Calculation Using Forces and Displacement
This problem involves calculating the work done on a particle when it is subjected to two constant forces and undergoes a displacement. The work done by a constant force is calculated using the dot product of the force vector and the displacement vector. When multiple forces act, we can find the resultant force first and then calculate the work done by the resultant force.
Step 1: Determine the Resultant Force
We are given two constant forces, \(\rm \vec P\) and \(\rm \vec Q\). The resultant force, \(\rm \vec F_{net}\), is the vector sum of these two forces.
- Force \(\rm \vec P = 2\hat{i} - 5\hat{j} + 6\hat{k}\)
- Force \(\rm \vec Q = -\hat{i} + 2\hat{j} - \hat{k}\)
To find the resultant force, we add the corresponding components of \(\rm \vec P\) and \(\rm \vec Q\):
\(\rm \vec F_{net} = \vec P + \vec Q\)
\(\rm \vec F_{net} = (2\hat{i} - 5\hat{j} + 6\hat{k}) + (-\hat{i} + 2\hat{j} - \hat{k})\)
\(\rm \vec F_{net} = (2 - 1)\hat{i} + (-5 + 2)\hat{j} + (6 - 1)\hat{k}\)
\(\rm \vec F_{net} = 1\hat{i} - 3\hat{j} + 5\hat{k}\)
Step 2: Calculate the Displacement Vector
The particle is displaced from point A to point B. The displacement vector, \(\rm \vec d\), is found by subtracting the position vector of the initial point (A) from the position vector of the final point (B).
- Initial position vector \(\rm \vec r_A = 4\hat{i} - 3\hat{j} - 2\hat{k}\)
- Final position vector \(\rm \vec r_B = 6\hat{i} + \hat{j} - 3\hat{k}\)
The displacement vector is calculated as:
\(\rm \vec d = \vec r_B - \vec r_A\)
\(\rm \vec d = (6\hat{i} + \hat{j} - 3\hat{k}) - (4\hat{i} - 3\hat{j} - 2\hat{k})\)
\(\rm \vec d = (6 - 4)\hat{i} + (1 - (-3))\hat{j} + (-3 - (-2))\hat{k}\)
\(\rm \vec d = 2\hat{i} + (1 + 3)\hat{j} + (-3 + 2)\hat{k}\)
\(\rm \vec d = 2\hat{i} + 4\hat{j} - \hat{k}\)
Step 3: Calculate the Work Done
The work done (\(W\)) is the dot product of the resultant force vector (\(\rm \vec F_{net}\)) and the displacement vector (\(\rm \vec d\)).
\(W = \vec F_{net} \cdot \vec d\)
\(W = (1\hat{i} - 3\hat{j} + 5\hat{k}) \cdot (2\hat{i} + 4\hat{j} - \hat{k})\)
To find the dot product, we multiply the corresponding components and sum the results:
\(W = (1 \times 2) + (-3 \times 4) + (5 \times -1)\)
\(W = 2 - 12 - 5\)
\(W = -10 - 5\)
\(W = -15\) units
Therefore, the work done when the particle is displaced from A to B is -15 units.
Important Questions from Applications of Vectors
A spacecraft located at î + 2ĵ + 3k̂ is subjected to a force λ k̂ by firing a rocket. The spacecraft is subjected to a moment of magnitude
A force of 78 grams acts at the point (2, 3, 5), the direction ratios of the line of action being 2, 2, 1. The magnitude of its moment about the line joining the origin to the point (12, 3, 4) is:
Let \(\rm \vec{a}\), \(\rm \vec{b}\) and \(\rm \vec{c}\) be the position vectors of the three vertices A, B, C of a triangle respectively. Then the area of this triangle is given by:
Forces 3î + 2ĵ + 5k̂ and 2î + ĵ - 3k̂ are acting on a particle and displace it from the point 2î - ĵ - 3k̂ to the point 4î - 3ĵ + 7k̂. The work done by the force is:
If \(\overrightarrow{AC}=2\hat{i}+\hat{j}+\hat{k}\) and \(\overrightarrow{BD}=-\hat{i}+3\hat{j}+2\hat{k}\) then the area of the quadrilateral ABCD is
