Question

A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.80 J/m2 )x2-(3.60 J/m3 )y3.

Part A. What is the magnitude of the acceleration of the block when it is at the point x= 0.38 m, y= 0.53 m?

Part B. What is the direction of the acceleration of the block when it is at the point x= 0.38 m , y= 0.53 m ?

Answer 1

x-component of force:
(Fx) = -∂U/∂x
(Fx) = -2(5.80)x
(Fx) = -11.60x

x-component of acceleration:
m(ax) = -11.60x
(0.0400 kg)(ax) = -11.60(0.38m)
(ax) = - 110 .2 m/s²

y-component of force:
(Fy) = -∂U/∂y
(Fy) = 3(3.60)y²
(Fy) = 10.80y²

y-component of acceleration:
m(ay) = 10.80y²
(0.0400 kg)(ay) = 10.80(0.53m)²
(ay) = 136.48m/s²

a = sqrt[(ax)² + (ay)²]
a = sqrt[(-110.2)² + (136.48)²]
a = 175.4 m/s²

tanθ = (ay)/(ax)
tanθ = (136.48)/(-110.2) (quadrant II)
θ = 128.92