Question
State the formula for calculating work done by a force. Are there any conditions or limitations in using it directly? If so, state those clearly. Is there any mathematical way out for it? Explain.
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Answer
1. Suppose a constant force $\vec{F}$ acting on a body produces a displacement $\vec{S}$ in the body along the positive X-direction. Then the work done by the force is given as, $W =$ F.s $\cos \theta$
Where $\theta$ is the angle between the applied force and displacement.
2. If displacement is in the direction of the force applied, $\theta=0^{\circ}$
$W =\overrightarrow{ F } \cdot \overrightarrow{ s }$
Conditions/limitations for application of work formula:
1. The formula for work done is applicable only if both force $\vec{F}$ and displacement $\vec{s}$ are constant and finite i.e., it cannot be applied when the force is variable.
2. The formula is not applicable in several real- life situations like lifting an object through several thousand kilometres since the gravitational force is not constant. It is not applicable to viscous forces like fluid resistance as they depend upon speed and thus are often not constant with time.
3. The method of integration has to be applied to find the work done by a variable force.
Integral method to find work done by a variable force:
Where $\theta$ is the angle between the applied force and displacement.
2. If displacement is in the direction of the force applied, $\theta=0^{\circ}$
$W =\overrightarrow{ F } \cdot \overrightarrow{ s }$
Conditions/limitations for application of work formula:
1. The formula for work done is applicable only if both force $\vec{F}$ and displacement $\vec{s}$ are constant and finite i.e., it cannot be applied when the force is variable.
2. The formula is not applicable in several real- life situations like lifting an object through several thousand kilometres since the gravitational force is not constant. It is not applicable to viscous forces like fluid resistance as they depend upon speed and thus are often not constant with time.
3. The method of integration has to be applied to find the work done by a variable force.
Integral method to find work done by a variable force:
- Let the force vary non-linearly in magnitude between the points A and B as shown in figure (a).
- In order to calculate the total work done during the displacement from $s_1$ to $s_2,$ we need to use integration. For integration, we need to divide the displacement into large numbers of infinitesimal (infinitely small) displacements.
- Let at $P_1$, the magnitude of force be $F = P_1P_1‘$. Due to this force, the body displaces through infinitesimally small displacement ds, in the direction of force.
It moves from $P_1$ to $P_2.$
$\therefore d \overrightarrow{ s }=\overrightarrow{ P _1 P _2}$
4. But direction of force and displacement are same, we have
$d \overrightarrow{ s }= P _1^{\prime} P _2^{\prime}$
5. $d \overrightarrow{ s }$ is so small that the force $F$ is practically constant for the displacement. As the force is constant, the area of the strip $\vec{F} \cdot d \vec{s}$ is the work done dW for this displacement.
6. Hence, small work done between $P_1$ to $P_2$ is $d W$ and is given by
$dW =\overrightarrow{ F } \cdot d \overrightarrow{ s }= P _1 P _1^{\prime} \times P _1^{\prime} P _2^{\prime}$
$\text { = Area of the strip } P _1 P _2 P _2^{\prime} P _1^{\prime} \text {. }$
The total work done can be found out by dividing the portion AB into small strips like $P_1P_2P_2‘P_1‘$ and taking sum of all the areas of the strips.
$∴ W = \int_{s_{1}}^{s 2} \vec{F} \cdot d \vec{s}=\text { Area } A B B^{\prime} A^{\prime}$
8. Method of integration is applicable if the exact way of variation in $\overrightarrow{ F }$ and $\overrightarrow{ s }$ is known and that function is integrable.
9. The work done by the non-linear variable force is represented by the area under the portion of force-displacement graph.
10. Similarly, in case of a linear variable force, the area under the curve from $s_1$ to $s_2$ (trapezium APQB) gives total work done W [figure (b)].
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