Question

The potential energy for a force field F is given by U[x,y]=sin[x+y]. The force acting on the particle of mass m at [0,pi/4] is
a] 1
b] root 2
c] 1/root 2
d] 0

Answer 1

Answer: a] 1unit

As we know that if the potential energy function is given to us that is U(x, y) then my differentiating it we can obtain the equation for force at any position also.

F= - dU/dx

This means that, if the potential energy vs position graph is given then the force can be obtain by the negative of the slope of that function at any point.

Here in the question, U(x, y) = sin[x+y]

F = - dU/dx i^ - dU/dy j^

F = cos(x+y) i^+ cos(x+y) j^

we have to find force acting on the particle at [0,π/4]

therefore, F(0,π/4) = cos(0+π/4)i^+ Cos(0+π/4)j^

F (0,π/4) = 1/√2(i^)+ 1/√2(j^)

|F| = √(( 1/√2) ^2+( 1/√2) ^2)

|F| = 1N

Therefore, Force acting on the particle of mass m at [0,π/4] is 1 N.