(A) The force components are derived from the negative gradient of the potential energy: $F_x = -\frac{\partial U}{\partial x} = -a$ $F_y = -\frac{\pa

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(A) The force components are derived from the negative gradient of the potential energy: $F_x = -\frac{\partial U}{\partial x} = -a$ $F_y = -\frac{\partial U}{\partial y} = -b$ The magnitude of the net force $F$ is given by: $F = \sqrt{F_x^2 + F_y^2} = \sqrt{(-a)^2 + (-b)^2} = \sqrt{a^2 + b^2}$ Using Newton's second law,the acceleration $a_{acc}$ is: $a_{acc} = \frac{F}{m} = \frac{\sqrt{a^2 + b^2}}{m} = \frac{(a^2 + b^2)^{1/2}}{m}$

Class 11 Physics Work, Energy and Power Potential Energy

Difficult

The potential energy of a body of mass $m$ is given by $U = ax + by$,where $x$ and $y$ are the position coordinates of the particle. The acceleration of the particle is:

A
$\frac{(a^2 + b^2)^{1/2}}{m}$
B
$\frac{a^2 + b^2}{m}$
C
$\frac{(a + b)^{1/2}}{m}$
D
$\frac{a + b}{m}$

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Class 11

Physics Questions

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Work, Energy and Power

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Potential Energy

$A$ particle moving along the $X$-axis has potential energy $U = 2 - 20x + 5x^2 \text{ J}$. The particle is released at $x = -3 \text{ m}$. What is the maximum value of $x$ reached by the particle (in $\text{ m}$)? ($x$ is in meters and $U$ is in joules)

MediumAP EAMCET 2020

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$A$ uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(1/n)^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface,the work done should be

DifficultJEE MAIN 2019

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$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity, the work required to pull the hanging part onto the table is

DifficultIIT 1985

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The potential energy between two atoms in a molecule is given by $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$,where $a$ and $b$ are positive constants and $x$ is the distance between the atoms. The atom is in stable equilibrium when:

DifficultAIPMT 1995

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When an arrow is released from a bow,from where does the arrow acquire its kinetic energy?

Medium

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The potential energy of a body of mass $m$ is given by $U = ax + by$,where $x$ and $y$ are the position coordinates of the particle. The acceleration of the particle is:

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$A$ particle moving along the $X$-axis has potential energy $U = 2 - 20x + 5x^2 \text{ J}$. The particle is released at $x = -3 \text{ m}$. What is the maximum value of $x$ reached by the particle (in $\text{ m}$)? ($x$ is in meters and $U$ is in joules)

$A$ uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(1/n)^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface,the work done should be

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity, the work required to pull the hanging part onto the table is

The potential energy between two atoms in a molecule is given by $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$,where $a$ and $b$ are positive constants and $x$ is the distance between the atoms. The atom is in stable equilibrium when:

When an arrow is released from a bow,from where does the arrow acquire its kinetic energy?

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