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Question: Find the magnitude of the force on the particle of mass m at (0, \(\dfrac{\pi }{4}\)), when the pote...

Find the magnitude of the force on the particle of mass m at (0, π4\dfrac{\pi }{4}), when the potential energy for a force field F is given by U (x, y) = sin⁡(x+y)\sin (x + y)
A. 1
B. 2\sqrt 2
C. 12\dfrac{1}{{\sqrt 2 }}
D. 0

Explanation

Solution

If potential is given as a function of position then force can be found by using the formula:
F⃗\vec F= −dU(x)dx\dfrac{{dU(x)}}{{dx}}
But here potential is a function of both x and y therefore we will find force by using formula:
F⃗=δU(x,y)δx+δU(x,y)δy\vec F = \dfrac{{\delta U(x,y)}}{{\delta x}} + \dfrac{{\delta U(x,y)}}{{\delta y}}
And after finding the value of F⃗\vec F , then we can find the magnitude by taking the modulus of F⃗\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.
F⃗\vec F= −dU(x)dx\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:
F⃗x\vec Fx = −δU(x,y)δx - \dfrac{{\delta U(x,y)}}{{\delta x}} (here function is differentiated w.r.t x, and y is treated as constant)
F⃗y=−δU(x,y)δy\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,
F⃗x\vec Fx = −δU(x,y)δx - \dfrac{{\delta U(x,y)}}{{\delta x}} =−δsin⁡(x+y)δx = - \dfrac{{\delta \sin (x + y)}}{{\delta x}} =−δsin⁡(x+y)δx×δ(x+y)δx = - \dfrac{{\delta \sin (x + y)}}{{\delta x}} \times \dfrac{{\delta (x + y)}}{{\delta x}} (derivative by chain rule)
=−cos⁡(x+y)×1= - \cos (x + y) \times 1 (∵dsin⁡xdx=cos⁡x)(\because \dfrac{{d\sin x}}{{dx}} = \cos x)
=−cos⁡(x+y)i^= - \cos (x + y)\hat i
Similarly,
F⃗y=−δU(x+y)δy=−δsin⁡(x+y)δy=−cos⁡(x+y)j^\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, π4\dfrac{\pi }{4}) this means x = 0 and y = π4\dfrac{\pi }{4} , putting these coordinates in the value of F⃗x\vec Fx and F⃗y\vec Fywe get:
F⃗x\vec Fx =−cos⁡(0+π4)i^=−cos⁡π4i^=−12i^ = - \cos (0 + \dfrac{\pi }{4})\hat i = - \cos \dfrac{\pi }{4}\hat i = - \dfrac{1}{{\sqrt 2 }}\hat i
And F⃗y\vec Fy =−cos⁡(0+π4)i^=−cos⁡π4i^=−12i^ = - \cos (0 + \dfrac{\pi }{4})\hat i = - \cos \dfrac{\pi }{4}\hat i = - \dfrac{1}{{\sqrt 2 }}\hat i
The magnitude of any vector z⃗=a^+b^\vec z = \hat a + \hat b is given by a2+b2\sqrt {{a^2} + {b^2}}
So here, F⃗=F⃗x+F⃗y=−12i^−12j^\vec F = \vec Fx + \vec Fy = - \dfrac{1}{{\sqrt 2 }}\hat i - \dfrac{1}{{\sqrt 2 }}\hat j
∴∣F⃗∣=(−12)2+(−12)2=12+12=1=1\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,F⃗x\vec Fx= −dU(x)dx\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:
x2=∣x∣\sqrt {{x^2}} = \left| x \right| not ±x \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.