The relationship between force F and displace | vedclass

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(B) The work done by a variable force is equal to the area under the force-displacement $(F-x)$ graph.For the interval $x = 1 \ m$ to $x = 5 \ m$,we c

Vedclass · Accepted Answer

$15$

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(B) The work done by a variable force is equal to the area under the force-displacement $(F-x)$ graph.For the interval $x = 1 \ m$ to $x = 5 \ m$,we calculate the area under the curve:$1$. From $x = 1 \ m$ to $x = 2 \ m$: The force is constant at $F = 10 \ N$. Area $= (2 - 1) 	imes 10 = 10 \ J$.$2$. From $x = 2 \ m$ to $x = 3 \ m$: The force is constant at $F = 5 \ N$. Area $= (3 - 2) 	imes 5 = 5 \ J$.$3$. From $x = 3 \ m$ to $x = 4 \ m$: The force is constant at $F = -5 \ N$. Area $= (4 - 3) 	imes (-5) = -5 \ J$.$4$. From $x = 4 \ m$ to $x = 5 \ m$: The force increases linearly from $0 \ N$ to $10 \ N$. Area $= \frac{1}{2} 	imes (5 - 4) 	imes 10 = 5 \ J$.Total work done $= 10 + 5 - 5 + 5 = 15 \ J$.

The relationship between force $F$ and displacement $x$ is shown in the figure. The work done by the object for a displacement from $x = 1 \ m$ to $x = 5 \ m$ is equal to ... $J$.

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