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(D) The change in potential energy $\Delta U$ is related to the work done by a conservative force $W_c$ as $\Delta U = -W_c$.Since $W_c = \int_{0}^{x_

Vedclass · Accepted Answer

$2, 1, 3$

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(D) The change in potential energy $\Delta U$ is related to the work done by a conservative force $W_c$ as $\Delta U = -W_c$.Since $W_c = \int_{0}^{x_1} F(x) dx$,the change in potential energy is $\Delta U = -\int_{0}^{x_1} F(x) dx$.This means $\Delta U$ is the negative of the area under the $F-x$ graph.For graph $1$,the force is positive,so the area is positive,and $\Delta U_1 = -(\frac{1}{2} F_1 x_1) = -0.5 F_1 x_1$.For graph $2$,the force is positive,so the area is positive,and $\Delta U_2 = -(F_1 x_1) = -1.0 F_1 x_1$.For graph $3$,the force is negative,so the area is negative,and $\Delta U_3 = -(-\frac{1}{2} F_1 x_1) = +0.5 F_1 x_1$.Comparing the values: $-1.0 F_1 x_1 Thus,the order from least to greatest is $2, 1, 3$.

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