Two forces F̅1 = î - ĵ + k̂ and F̅2 = 4î + 2ĵ + 3k̂ act on a particle and displace it from the point (0, 1, 2) to (1, -2, 3), then the total work done is
6 unit
Calculating Total Work Done by Forces and Displacement
When one or more forces act on a particle and cause a displacement, the work done by the forces is calculated using the dot product of the resultant force and the displacement vector. This is a fundamental concept in physics problem solving related to energy and mechanics.
Understanding the Given Forces and Displacement
We are given two forces acting on a particle:
- Force 1: $\vec{F_1} = \hat{i} - \hat{j} + \hat{k}$
- Force 2: $\vec{F_2} = 4\hat{i} + 2\hat{j} + 3\hat{k}$
The particle is displaced from an initial point A to a final point B:
- Initial Position A: (0, 1, 2)
- Final Position B: (1, -2, 3)
Finding the Resultant Force
The total or resultant force acting on the particle is the vector sum of the individual forces.
Let the total force be $\vec{F}$. Then,
$$ \vec{F} = \vec{F_1} + \vec{F_2} $$ $$ \vec{F} = (\hat{i} - \hat{j} + \hat{k}) + (4\hat{i} + 2\hat{j} + 3\hat{k}) $$To perform vector addition, we add the corresponding components:
$$ \vec{F} = (1+4)\hat{i} + (-1+2)\hat{j} + (1+3)\hat{k} $$ $$ \vec{F} = 5\hat{i} + \hat{j} + 4\hat{k} $$Determining the Displacement Vector
The displacement vector $\vec{d}$ is found by subtracting the initial position vector from the final position vector.
The position vector for point A (0, 1, 2) is $\vec{r_A} = 0\hat{i} + 1\hat{j} + 2\hat{k}$.
The position vector for point B (1, -2, 3) is $\vec{r_B} = 1\hat{i} - 2\hat{j} + 3\hat{k}$.
The displacement vector is:
$$ \vec{d} = \vec{r_B} - \vec{r_A} $$ $$ \vec{d} = (1\hat{i} - 2\hat{j} + 3\hat{k}) - (0\hat{i} + 1\hat{j} + 2\hat{k}) $$ $$ \vec{d} = (1-0)\hat{i} + (-2-1)\hat{j} + (3-2)\hat{k} $$ $$ \vec{d} = \hat{i} - 3\hat{j} + \hat{k} $$Calculating Work Done Using the Dot Product
The work done $W$ by the total force $\vec{F}$ during the displacement $\vec{d}$ is given by the dot product of these two vectors:
$$ W = \vec{F} \cdot \vec{d} $$We have $\vec{F} = 5\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{d} = \hat{i} - 3\hat{j} + \hat{k}$.
To perform the dot product, we multiply the corresponding components and sum the results:
$$ W = (5)(1) + (1)(-3) + (4)(1) $$ $$ W = 5 - 3 + 4 $$ $$ W = 2 + 4 $$ $$ W = 6 $$Therefore, the total work done is 6 units.
Final Result for Total Work Done
After calculating work done from the given forces and displacement, we find the total work done is 6 units. This calculation involves finding the resultant force through vector addition and then applying the dot product formula for work done in vector form. This type of physics problem is common when dealing with forces and motion in multiple dimensions.
Important Questions from Scalar and Vector Product
If \(\overrightarrow a \) and \(\overrightarrow b\) are two unit vectors inclined to x - axis at angles 30° and 120°, then \(\left| {\overrightarrow a + \overrightarrow b } \right|\) equals
The value of \(\left( {\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right)} \right]\) is:
The work done in moving an object along a vector \(\widehat d = 3\widehat i + 2\widehat j - 5\widehat k\) (if the applied force is \(\overrightarrow F = 2\widehat i - \widehat j - \widehat k\)) is
If \(\bar a\) and \(\bar b\) are unit vectors and θ is the angle between them then \(\left| {\frac{{\bar a - \bar b}}{2}} \right|\) is
If \(\vec{A}\) and \(\vec{B}\) are two vectors, then angle between vectors \(\vec{A}+\vec{B}\) and \(\vec{A}\times\vec{B}\) is
