The relationship between force and position i | vedclass

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(A) The work done by a variable force is equal to the area under the force-position $(F-x)$ graph.To find the work done from $x = 1 \, cm$ to $x = 5 \

Vedclass · Accepted Answer

$20$

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(A) The work done by a variable force is equal to the area under the force-position $(F-x)$ graph.To find the work done from $x = 1 \, cm$ to $x = 5 \, cm$,we calculate the area under the curve between these limits:$1$. From $x = 1 \, cm$ to $x = 2 \, cm$: The force is $10 \, dyne$. Area = $10 	imes (2 - 1) = 10 \, erg$.$2$. From $x = 2 \, cm$ to $x = 3 \, cm$: The force is $20 \, dyne$. Area = $20 	imes (3 - 2) = 20 \, erg$.$3$. From $x = 3 \, cm$ to $x = 4 \, cm$: The force is $-20 \, dyne$. Area = $-20 	imes (4 - 3) = -20 \, erg$.$4$. From $x = 4 \, cm$ to $x = 5 \, cm$: The force is $10 \, dyne$. Area = $10 	imes (5 - 4) = 10 \, erg$.Total work done $W = 10 + 20 - 20 + 10 = 20 \, erg$.

The relationship between force and position is shown in the figure given (in one-dimensional case). The work done by the force in displacing a body from $x = 1 \, cm$ to $x = 5 \, cm$ is .......... $ergs$.

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