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Standard XII

Physics

Conservative Force as Gradient of Potential

A particle ha...

Question

A particle has potential energy dependent on its position on the $x$ axis, represented by the function $U (x) = e^{2 x} + 1$ for all real values of $x$ , where $U (x)$ and $x$ are given in standard units. The force it feels at position $x = 1$ is closest to
[Take $e = 2.72$ ]

8.39 N

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- 8.39 N

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14.8 N

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- 14.8 N

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Solution

The correct option is D $- 14.8 N$
The potential energy of the particle $U (x) = e^{2 x} + 1$
Thus force acting on the particle $F = - \frac{d U (x)}{d x} = - 2 e^{2 x} N$
$\Rightarrow {\begin{matrix} F \end{matrix} |}_{x = 1} = - 2 e^{2} = - 14.8 N$

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A particle has potential energy dependent on its position on the x axis, represented by the function U(x)=e2x+1 for all real values of x, where U(x) and x are given in standard units. The force it feels at position x=1 is closest to [Take e=2.72]

The correct option is D −14.8 NThe potential energy of the particle U(x)=e2x+1 Thus force acting on the particle F=−dU(x)dx=−2e2x N ⇒F|x=1=−2e2=−14.8 N

A particle has potential energy dependent on its position on the $x$ axis, represented by the function $U (x) = e^{2 x} + 1$ for all real values of $x$ , where $U (x)$ and $x$ are given in standard units. The force it feels at position $x = 1$ is closest to
[Take $e = 2.72$ ]

The correct option is D $- 14.8 N$
The potential energy of the particle $U (x) = e^{2 x} + 1$
Thus force acting on the particle $F = - \frac{d U (x)}{d x} = - 2 e^{2 x} N$
$\Rightarrow {\begin{matrix} F \end{matrix} |}_{x = 1} = - 2 e^{2} = - 14.8 N$