The acceleration-displacement graph of a 10 | vedclass

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(A) The work done $W$ is given by the integral of force with respect to displacement: $W = \int F \, dx = \int m \cdot a \, dx = m \int a \, dx$.Here,

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$8 	imes 10^{-2} \ J$

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(A) The work done $W$ is given by the integral of force with respect to displacement: $W = \int F \, dx = \int m \cdot a \, dx = m \int a \, dx$.Here,$\int a \, dx$ represents the area under the acceleration-displacement graph.The area of the triangle formed by the graph is $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Base $= 8 \ cm = 8 	imes 10^{-2} \ m$ and Height $= 20 \ cm/s^2 = 20 	imes 10^{-2} \ m/s^2$.$	ext{Area} = \frac{1}{2} 	imes (8 	imes 10^{-2} \ m) 	imes (20 	imes 10^{-2} \ m/s^2) = 80 	imes 10^{-4} \ m^2/s^2 = 8 	imes 10^{-3} \ m^2/s^2$.Now,$W = m 	imes 	ext{Area} = 10 \ kg 	imes (8 	imes 10^{-3} \ m^2/s^2) = 80 	imes 10^{-3} \ J = 8 	imes 10^{-2} \ J$.

The acceleration-displacement graph of a $10 \ kg$ particle is given. Find the work done as the particle moves from $x = 0 \ cm$ to $x = 8 \ cm$.

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