QUESTION 9
Easy
The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work \(w\) and distance \(d\)), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.
SOLUTION
1
Express the relationship as a linear equation
Work done(\(w\)) = constant force(\(F\)) × distance travelled in direction of force(\(d\))
⇒ \(w = F . d\)
Given,\( F = 3 \) units(constant force)
Thus \(w = 3d\) is the linear equation in two variables(\(w\) and \(d\)).
⇒ \(w = F . d\)
Given,\( F = 3 \) units(constant force)
Thus \(w = 3d\) is the linear equation in two variables(\(w\) and \(d\)).
2
Find the points for drawing the graph
When \(d = 0\), \(w = 3(0) = 0\). So the point is \((0,0)\)
When \(d = 1\), \(w = 3(1) = 3\). So the point is \((1,3)\)
When \(d = 2\), \(w = 3(2) = 6\). So the point is \((2,6)\)
When \(d = 3\), \(w = 3(3) = 9\). So the point is \((3,9)\)
When \(d = 1\), \(w = 3(1) = 3\). So the point is \((1,3)\)
When \(d = 2\), \(w = 3(2) = 6\). So the point is \((2,6)\)
When \(d = 3\), \(w = 3(3) = 9\). So the point is \((3,9)\)
3
Find work done when \(d = 2\)
Substituting \(d = 2\) into \(w = 3d\), we get
\(w = 3 × 2 = 6\)
Therefore, work done = \(6\) units.
\(w = 3 × 2 = 6\)
Therefore, work done = \(6\) units.
4
Verify by plotting on the graph
On the graph, when \(d = 2\), \(w = 6\), so the point lies on the line, confirming the work done when \(d = 2\) is \(6\) units.
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Final Answer : later
Concept Note
For a constant force, \(w = F × d\) is a linear equation in two variables. To find work for any distance, substitute into \(w = F × d\).
The graph is a straight line through the origin with slope equal to \(F\).