The potential energy for a force field #vecF# is given by #U(x,y) = sin(x+y)#. The magnitude of force acting on the particle of mass m at #(0,pi/4)# is?

A) 1
B) #sqrt2#
C) #1/sqrt2#
D) #0#

Question

A) 1
B) #sqrt2#
C) #1/sqrt2#
D) #0#

Benjamin Asiedu · Accepted Answer

#vec F(x,y) = - \grad U(x,y)#
#F(x,y) = |\gradU(x,y)|; \qquad F(0,pi/4) = 1# #vec F(x,y) = -\grad U(x,y)# #\grad U(x,y)= \frac{\delU}{\del x} hat x + \frac{\delU}{\dely} hat y = cos(x+y) (hatx+haty)# #F(x,y) = |\gradU(x,y)|# #F(0,pi/4) = |\gradU(0,pi/4)| = |cos(pi/4)||hatx+haty| # #F(0,pi/4)= 1/sqrt{2}\sqrt{1^2+1^2} = 1 "unit"#

Elena Albarran · Answer

undefined To find the magnitude of the force acting on a particle at a given point in a force field, you can use the relationship between force and potential energy. The force $ \vec{F} $ acting on the particle is the negative gradient of the potential energy function $ U(x, y) $, given by $ \vec{F} = -\nabla U $, where $ \nabla $ is the gradient operator.

The gradient of the potential energy function $ U(x, y) $ is given by:

$$ \nabla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} \right) $$

Given that $ U(x, y) = \sin(x + y) $, we need to compute the partial derivatives:

$$ \frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(\sin(x + y)) = \cos(x + y) $$

$$ \frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(\sin(x + y)) = \cos(x + y) $$

So, the force $ \vec{F} $ acting on the particle at any point $(x, y)$ is:

$$ \vec{F} = -\left( \cos(x + y), \cos(x + y) \right) $$

Now, to find the magnitude of the force at the point $(0, \frac{\pi}{4})$, substitute $x = 0$ and $y = \frac{\pi}{4}$ into the expression for $ \vec{F} $:

$$ \vec{F} = -\left( \cos(0 + \frac{\pi}{4}), \cos(0 + \frac{\pi}{4}) \right) = -\left( \cos(\frac{\pi}{4}), \cos(\frac{\pi}{4}) \right) = -\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$

The magnitude of the force is the square root of the sum of the squares of its components:

$$ |\vec{F}| = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1 $$

So, the magnitude of the force acting on the particle at the point $(0, \frac{\pi}{4})$ is $1$.

The potential energy for a force field #vecF# is given by #U(x,y) = sin(x+y)#. The magnitude of force acting on the particle of mass m at #(0,pi/4)# is?

A) 1 B) #sqrt2# C) #1/sqrt2# D) #0#

A) 1
B) #sqrt2#
C) #1/sqrt2#
D) #0#