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Question

Physics Question on work, energy and power

The graphs below show the magnitude of the force on a particle as it moves along the positive XX -axis from the origin to XX1X - X_{1} . The force is parallel to the XX -axis and conservative. The maximum magnitude F1F_{1} has the same value for all graphs. Rank the situations according to the change in the potential energy associated with the force, least (or most negative) to greatest (or most positive).

A

(i), (ii), (iii)

B

(i) (iii), (ii),

C

(iii), (ii), (i)

D

(ii), (i), (iii)

Answer

(ii), (i), (iii)

Explanation

Solution

As we know, for a conservative force field
dU=FcondrdU=-F_{con} dr
where, dUdU = change in potential energy,
FconF_{con}= conservative force (or FinF_{in})
and drdr = change in position of the particle
dU=FindrdU =-F_{in} dr or ΔU=r1r2Findr\Delta U=-\int\limits_{r_1}^{r_{2}} F_{in} dr
U2U1=r1r2FindrU_{2}-U_{1}=-\int\limits_{r_1}^{r_{2}} F_{in} dr =-work done by FinF_{in} (or Win)W_{in})
For graph (i)\left(i\right), Win=F1x12W_{in}=\frac{F_{1}\cdot x_{1}}{2}
For graph (ii)\left(ii\right), Win=F1x1W_{in}=F_{1}\cdot x_{1}
and for graph (iii)\left(iii\right), Win=F1x12W_{in}=\frac{-F_{1}\cdot x_{1}}{2}
Thus, change in potential energy
For graph (i)\left(i\right), ΔU1=F1x12\Delta U_{1}=\frac{-F_{1}x_{1}}{2} graph (ii)\left(ii\right) ΔU2=F1x1\Delta U_{2}=-F_{1}x_{1}
graph, (iii)\left(iii\right) ΔU3=F1x12\Delta U_{3}=\frac{F_{1}x_{1}}{2}
Thus, we have,
$\Delta U_{2} < , \Delta U_{1}