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(A) The work done by a force is equal to the area under the force-displacement graph.We need to calculate the area under the graph from $x = 1 \; cm$

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The work done by a force is equal to the area under the force-displacement graph.We need to calculate the area under the graph from $x = 1 \; cm$ to $x = 5 \; cm$.$1$. From $x = 1$ to $x = 2$,the force is $10 \; dynes$. Area $A_1 = 10 	imes (2 - 1) = 10 \; erg$.$2$. From $x = 2$ to $x = 3$,the force is $20 \; dynes$. Area $A_2 = 20 	imes (3 - 2) = 20 \; erg$.$3$. From $x = 3$ to $x = 4$,the force is $-20 \; dynes$. Area $A_3 = -20 	imes (4 - 3) = -20 \; erg$.$4$. From $x = 4$ to $x = 5$,the force is $10 \; dynes$. Area $A_4 = 10 	imes (5 - 4) = 10 \; erg$.Total work done $W = A_1 + A_2 + A_3 + A_4 = 10 + 20 - 20 + 10 = 20 \; erg$.

The figure shows the relationship between force and position. What is the work done by the force for the displacement of the object from $x = 1 \; cm$ to $x = 5 \; cm$ in ergs?

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