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)= (5.90 J/m2 )x2-(3.50 J/m3 )y3.

A)

What is the magnitude of the acceleration of the block when it is at the point x= 0.39 m , y= 0.68 m ?

Express your answer with the appropriate units.

B) What is the direction of the acceleration of the block when it is at the point x= 0.39 m , y= 0.68 m ?

Answer 1

The relation between force ( F ) and the potential energy function ( U ) is given by

  

Where,

  

Given,

  

So,

    

Therefore, acceleration of the block is

(From Newton's second law of motion)

(A) At x = 0.39 m and y = 0.68 m , acceleration is

So, magnitude of acceleration at this point is

(B) The direction of the acceleration (angle with positive x-axis) at this point can be determined as

Hence, the acceleration vector is making an angle of -46.5^o  with positive x- axis in clockwise direction (in 4th quadrant) .  

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