A force \({\rm{\vec F}} = {\rm{\hat i}} + 3{\rm{\hat j}} + 2{\rm{\hat k}}\) acts on a particle to displace it from the point \({\rm{A}}\left( {{\rm{\hat i}} + 2{\rm{\hat j}} - 3{\rm{\hat k}}} \right)\) to the point \({\rm{B}}\left( {3{\rm{\hat i}} - {\rm{\hat j}} + 5{\rm{\hat k}}} \right)\) .The work done by the force will be
9 units
Calculating Work Done by a Force
Understanding how to calculate the work done by a constant force is a fundamental concept in physics. Work is defined as the product of the force applied and the displacement caused by that force in the direction of the force. Mathematically, for a constant force \(\vec F\) causing a displacement \(\vec d\), the work done \(W\) is given by the dot product of the force vector and the displacement vector:
\(W = \vec F \cdot \vec d\)
Identifying the Force and Displacement Vectors
In this problem, we are given the force vector \(\vec F\) and the initial and final positions of the particle. First, let's write down the given force vector:
Force vector, \({\rm{\vec F}} = {\rm{\hat i}} + 3{\rm{\hat j}} + 2{\rm{\hat k}}\)
Next, we need to find the displacement vector \(\vec d\). The displacement vector is the vector pointing from the initial position (point A) to the final position (point B). The position vector of point A is \({\rm{\vec r}}_{\rm{A}} = {\rm{\hat i}} + 2{\rm{\hat j}} - 3{\rm{\hat k}}\) and the position vector of point B is \({\rm{\vec r}}_{\rm{B}} = 3{\rm{\hat i}} - {\rm{\hat j}} + 5{\rm{\hat k}}\). The displacement vector \(\vec d\) is given by:
\(\vec d = {\rm{\vec r}}_{\rm{B}} - {\rm{\vec r}}_{\rm{A}}\)
Let's calculate the displacement vector:
\(\vec d = \left( {3{\rm{\hat i}} - {\rm{\hat j}} + 5{\rm{\hat k}}} \right) - \left( {{\rm{\hat i}} + 2{\rm{\hat j}} - 3{\rm{\hat k}}} \right)\)
\(\vec d = \left( {3 - 1} \right){\rm{\hat i}} + \left( { - 1 - 2} \right){\rm{\hat j}} + \left( {5 - \left( { - 3} \right)} \right){\rm{\hat k}}\)
\(\vec d = \left( 2 \right){\rm{\hat i}} + \left( { - 3} \right){\rm{\hat j}} + \left( {5 + 3} \right){\rm{\hat k}}\)
\(\vec d = 2{\rm{\hat i}} - 3{\rm{\hat j}} + 8{\rm{\hat k}}\)
Calculating the Work Done using Dot Product
Now that we have the force vector \(\vec F\) and the displacement vector \(\vec d\), we can calculate the work done \(W\) using their dot product:
\(W = \vec F \cdot \vec d\)
\(W = \left( {{\rm{\hat i}} + 3{\rm{\hat j}} + 2{\rm{\hat k}}} \right) \cdot \left( {2{\rm{\hat i}} - 3{\rm{\hat j}} + 8{\rm{\hat k}}} \right)\)
To compute the dot product of two vectors \(\vec A = A_x{\rm{\hat i}} + A_y{\rm{\hat j}} + A_z{\rm{\hat k}}\) and \(\vec B = B_x{\rm{\hat i}} + B_y{\rm{\hat j}} + B_z{\rm{\hat k}}\), we use the formula \(\vec A \cdot \vec B = A_x B_x + A_y B_y + A_z B_z\). Applying this to our force and displacement vectors:
\(W = \left( 1 \right)\left( 2 \right) + \left( 3 \right)\left( { - 3} \right) + \left( 2 \right)\left( 8 \right)\)
\(W = 2 - 9 + 16\)
\(W = -7 + 16\)
\(W = 9\)
The work done by the force on the particle is 9 units.
Summary of Work Done Calculation
| Quantity | Value | Unit |
|---|---|---|
| Force Vector (\(\vec F\)) | \({\rm{\hat i}} + 3{\rm{\hat j}} + 2{\rm{\hat k}}\) | (Implied) |
| Initial Position (A) | \({\rm{\hat i}} + 2{\rm{\hat j}} - 3{\rm{\hat k}}\) | (Implied) |
| Final Position (B) | \(3{\rm{\hat i}} - {\rm{\hat j}} + 5{\rm{\hat k}}\) | (Implied) |
| Displacement Vector (\(\vec d = \vec r_B - \vec r_A\)) | \(2{\rm{\hat i}} - 3{\rm{\hat j}} + 8{\rm{\hat k}}\) | (Implied) |
| Work Done (\(W = \vec F \cdot \vec d\)) | 9 | units |
Thus, the work done by the force \(\vec F = {\rm{\hat i}} + 3{\rm{\hat j}} + 2{\rm{\hat k}}\) in displacing the particle from point A to point B is 9 units.
Revision Table: Key Concepts for Work Done
| Concept | Description | Formula |
|---|---|---|
| Work Done | Energy transferred by a force acting over a distance. Scalar quantity. | \(W = F \cdot d \cdot \cos\theta\) or \(W = \vec F \cdot \vec d\) |
| Force Vector | Vector quantity representing the push or pull on an object. | \(\vec F\) |
| Displacement Vector | Vector quantity representing the change in position of an object. Points from initial to final position. | \(\vec d = \vec r_{\text{final}} - \vec r_{\text{initial}}\) |
| Dot Product | A scalar product of two vectors. Result is a scalar quantity. | \(\vec A \cdot \vec B = |\vec A| |\vec B| \cos\theta\) or \(\vec A \cdot \vec B = A_x B_x + A_y B_y + A_z B_z\) |
Additional Information: Work and Energy
Work done is a measure of energy transfer. When a force does positive work, it transfers energy to the object, often increasing its kinetic energy. When a force does negative work, it removes energy from the object, often decreasing its kinetic energy.
- Units of Work: In the International System of Units (SI), the unit of work is the Joule (J). One Joule is defined as the work done by a force of one Newton causing a displacement of one meter in the direction of the force (1 J = 1 N⋅m). The problem states "units", implying the answer is a scalar value derived from the given vector components.
- Work-Energy Theorem: This important theorem states that the net work done on an object by all forces is equal to the change in its kinetic energy (\(W_{\text{net}} = \Delta KE\)).
- Conservative vs. Non-Conservative Forces:
- Conservative forces (like gravity and elastic force) do work that is independent of the path taken; the work done depends only on the initial and final positions.
- Non-conservative forces (like friction and air resistance) do work that depends on the path taken.
Calculating work done by finding the dot product of the force and displacement vectors is a common method for constant forces. The result of 9 units represents the scalar amount of energy transferred by the force during the displacement.
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