First slide

Work done by Diff type of forces

Question

A particle of mass 2 kg is moved along a line $3 y = 4 x + 10$ under the influence of a force $\vec{F} = (4 \hat{i} - 3 \hat{j}) N$ . Find the work done by the force between the points $(- \frac{10}{4}, 0)$ and $(0, \frac{10}{3})$

Easy

A

20 J
B

30 J
C

36.5 J
D

Zero

Solution

The particle is moved along the line $3 y = 4 x + 10, y = \frac{4}{3} x + \frac{10}{3} = mx + c$ , . Hence the slope of this line is 4/3. Again for the force acting $slope = tanθ = \frac{F_{y}}{F_{x}} = - \frac{3}{4}$ . So product of the slopes= -1. We can say that the two lines are perpendicular to each other.
$W = F . S . Cosθ = F . S . Cos 90^{0} = 0$

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