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?

Question

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

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Answer 1 · 2017-11-11T17:55:51

#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"#