
A force \[\overrightarrow F = \left( {5\widehat i + 3\widehat j} \right)\]newton is applied over a particle which displaces it from its origin to the point \[\overrightarrow r = \left( {2\widehat i - \widehat j} \right)\] meters. Find the work done on the particle.
Answer
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Hint:Before we start addressing the problem, we need to know about the displacement of a particle. It is defined as the change in the position of an object. Since it is a vector quantity it has both direction and magnitude.
Formula Used:
Formula to find the work done is,
\[W = \overrightarrow F \cdot \overrightarrow S \]
Where, \[\overrightarrow F \] is force applied and \[\overrightarrow S \] is displacement.
Complete step by step solution:
Consider a particle on which a force is applied, when this force is applied the particle undergoes displacement that is the particle gets displaced from the origin. We need to find the work done in moving the particle from the start point to the endpoint. They have given the force \[\overrightarrow F = \left( {5\widehat i + 3\widehat j} \right)\] and the displacement \[\overrightarrow r = \left( {2\widehat i - \widehat j} \right) = \overrightarrow S \].
To find work done the formula is given by,
\[W = \overrightarrow F \cdot \overrightarrow S \]
On substituting the value of force and the displacement we get,
\[W = \left( {5\widehat i - 3\widehat j} \right) \cdot \left( {2\widehat i - \widehat j} \right)\]
\[\Rightarrow W = 10 - 3\]
\[\therefore W = 7J\]
Here, we have used the concept of properties of unit vectors i.e., \[\widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1\]\[\widehat i \cdot \widehat j = \widehat j \cdot \widehat k = \widehat k \cdot \widehat i = 0\]and the angle between the coordinate axes is \[{90^0}\] and \[\cos {90^0} = 0\]
Therefore, the work done on the particle is 7J.
Note:Basically the displacement tells the direct length between any two points when measured along the minimum path between them. Some of the examples of displacement are, a teacher walking across the black board, a jogger on a jogging track etc. The displacement of an object value can be positive, negative and even zero, but the distance can have only positive values and cannot be negative.
Formula Used:
Formula to find the work done is,
\[W = \overrightarrow F \cdot \overrightarrow S \]
Where, \[\overrightarrow F \] is force applied and \[\overrightarrow S \] is displacement.
Complete step by step solution:
Consider a particle on which a force is applied, when this force is applied the particle undergoes displacement that is the particle gets displaced from the origin. We need to find the work done in moving the particle from the start point to the endpoint. They have given the force \[\overrightarrow F = \left( {5\widehat i + 3\widehat j} \right)\] and the displacement \[\overrightarrow r = \left( {2\widehat i - \widehat j} \right) = \overrightarrow S \].
To find work done the formula is given by,
\[W = \overrightarrow F \cdot \overrightarrow S \]
On substituting the value of force and the displacement we get,
\[W = \left( {5\widehat i - 3\widehat j} \right) \cdot \left( {2\widehat i - \widehat j} \right)\]
\[\Rightarrow W = 10 - 3\]
\[\therefore W = 7J\]
Here, we have used the concept of properties of unit vectors i.e., \[\widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1\]\[\widehat i \cdot \widehat j = \widehat j \cdot \widehat k = \widehat k \cdot \widehat i = 0\]and the angle between the coordinate axes is \[{90^0}\] and \[\cos {90^0} = 0\]
Therefore, the work done on the particle is 7J.
Note:Basically the displacement tells the direct length between any two points when measured along the minimum path between them. Some of the examples of displacement are, a teacher walking across the black board, a jogger on a jogging track etc. The displacement of an object value can be positive, negative and even zero, but the distance can have only positive values and cannot be negative.
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