Exercise 7.33 A small block with mass 0.0400 kg is moving in the xy -plane. The

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Exercise 7.33

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)= (5.50 J/m2 )x2-(3.85 J/m3 )y3.

Part A

What is the magnitude of the acceleration of the block when it is at the point x= 0.35 m , y= 0.52 m ?

Express your answer with the appropriate units.

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Part B

What is the direction of the acceleration of the block when it is at the point x= 0.35 m , y= 0.52 m ?

Exercise 7.33

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)= (5.50 J/m2 )x2-(3.85 J/m3 )y3.

Part A

What is the magnitude of the acceleration of the block when it is at the point x= 0.35 m , y= 0.52 m ?

Express your answer with the appropriate units.

a=

SubmitMy AnswersGive Up

Part B

What is the direction of the acceleration of the block when it is at the point x= 0.35 m , y= 0.52 m ?

=   counterclockwise from the +x-axis

Explanation / Answer

U = 5.5*x^2 - 3.85y^3

So, F = -dU/dx i - dU/dy j

= -11*x i - 11.55*y^2 j

So, a = F/m = (-11*x i - 11.55*y^2 j )/0.04

= ( -275*x i - 288.8y^2 j ) m/s2

So, magnitude of acceleration at x = 0.35 , y = 0.52,

a =0 sqrt((275*0.35)^2 + (288.8*0.52^2)^2) = 123.9 m/s2 <------answer

b)

direction = 180 + atan((288.8*0.52^2)/((275*0.35))

= 219.1 deg

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