Question

Find the magnitude of the force on the particle of mass m at (0, $\dfrac{\pi }{4}$), when the potential energy for a force field F is given by U (x, y) = $\sin (x + y)$
A. 1
B. $\sqrt 2 $
C. $\dfrac{1}{{\sqrt 2 }}$
D. 0

Answer 1

Hint: If potential is given as a function of position then force can be found by using the formula:
$\vec F$= −$\dfrac{{dU(x)}}{{dx}}$
But here potential is a function of both x and y therefore we will find force by using formula:
$\vec F = \dfrac{{\delta U(x,y)}}{{\delta x}} + \dfrac{{\delta U(x,y)}}{{\delta y}}$
And after finding the value of $\vec F$ , then we can find the magnitude by taking the modulus of $\vec F$

Complete step by step answer:
When the potential energy of a particle is given as a function of U(x), then the force at any position can be calculated by taking the derivative of the potential.
$\vec F$= −$\dfrac{{dU(x)}}{{dx}}$
But here the potential is a function of both x and y, so we will find the x and y components of the force by taking partial derivative of the function U (x, y) such that:
$\vec Fx$ = $ - \dfrac{{\delta U(x,y)}}{{\delta x}}$ (here function is differentiated w.r.t x, and y is treated as constant)
$\vec Fy = - \dfrac{{\delta U(x,y)}}{{\delta y}}$ (here function is differentiated w.r.t y, and x is treated as constant)
Now here,
$\vec Fx$ = $ - \dfrac{{\delta U(x,y)}}{{\delta x}}$$ = - \dfrac{{\delta \sin (x + y)}}{{\delta x}}$$ = - \dfrac{{\delta \sin (x + y)}}{{\delta x}} \times \dfrac{{\delta (x + y)}}{{\delta x}}$ (derivative by chain rule)
        $ = - \cos (x + y) \times 1$ $(\because \dfrac{{d\sin x}}{{dx}} = \cos x)$
        $ = - \cos (x + y)\hat i$
Similarly,
\[\vec Fy = - \dfrac{{\delta U(x + y)}}{{\delta y}} = - \dfrac{{\delta \sin (x + y)}}{{\delta y}} = - \cos (x + y)\hat j\]
When the particle of mass m is at (0, $\dfrac{\pi }{4}$) this means x = 0 and y = $\dfrac{\pi }{4}$ , putting these coordinates in the value of $\vec Fx$ and $\vec Fy$we get:
$\vec Fx$ $ = - \cos (0 + \dfrac{\pi }{4})\hat i = - \cos \dfrac{\pi }{4}\hat i = - \dfrac{1}{{\sqrt 2 }}\hat i$
And $\vec Fy$$ = - \cos (0 + \dfrac{\pi }{4})\hat i = - \cos \dfrac{\pi }{4}\hat i = - \dfrac{1}{{\sqrt 2 }}\hat i$
The magnitude of any vector $\vec z = \hat a + \hat b$ is given by $\sqrt {{a^2} + {b^2}} $
So here, \[\vec F = \vec Fx + \vec Fy = - \dfrac{1}{{\sqrt 2 }}\hat i - \dfrac{1}{{\sqrt 2 }}\hat j\]
$\therefore \left| {\vec F} \right| = \sqrt {{{\left( { - \dfrac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( { - \dfrac{1}{{\sqrt 2 }}} \right)}^2}} = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}} = \sqrt 1 = 1$
Hence, the magnitude of force is 1.

So, the correct answer is “Option A”.

Additional Information:
Since,$\vec Fx$= −$\dfrac{{dU(x)}}{{dx}}$
 Graphically, this signifies that if we have potential energy vs. position graph then the force is the negative of the slope of the function for potential at that point.
F=−(slope)
Here, the minus sign signifies that if the slope is positive, the force is to the left and if slope is negative the force is to the right.

Note:
$\sqrt {{x^2}} = \left| x \right|$ not $ \pm x$. Hence the magnitude of any vector quantity is always positive, never negative, since it is just the length of the vector quantity i.e. distance of the terminal point from origin.