Question

A ball of mass 0.0400kg moves in the xy plane. The net force on the ball is described by the potential energy function U(x,y)=(5.80J/m2)x2 - (3.60J/m3)y3 Find the magnitude of the ball acceleration at the point (0.300m, 0.600m)

Answer 1

130 m/s²

Step-by-step explanation:

The x and y components of the force are the negative partial derivatives of the potential energy function with respect to x and y.

Fx = -∂U/∂x

Fy = -∂U/∂y

Given that U = 5.80x² − 3.60y³:

Fx = -11.6x

Fy = 10.8y²

At the point (0.300, 0.600):

Fx = -11.6 (0.300) = -3.480

Fy = 10.8 (0.600)² = 3.888

The magnitude of the net force is:

F = √((-3.480)² + (3.888)²)

F = 5.218 N

The acceleration is therefore:

a = F/m

a = 5.218 N / 0.0400 kg

a = 130 m/s²