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

Question

Exercise 7.33 A small block with mass 0.0400, is moving in the xy-plane. The net force on the block is described by the potential- energy function(5.90) -(3.65 Part A What is the magnitude of the acceleration of the block when it is at the point0.400.60 Express your answer with the appropriate units. Value Units Submit My Answers Give Up Part B What is the direction of the acceleration of the block when it is at the point0.40 m0.60 ? counterclockwise from the +r-axis Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

U(x,y) = (5.90 J/m^2)x^2 - (3.65 J/m^3)y^3

F = (-dU/dx , -dU/dy)

F = ((-5.90 J/m^2)2x , (3.65 J/m^3)3y^2)

F = (-11.8x J/m^2, 10.95 y^2 J/m^3)

(-11.8x J/m^2, 10.95 y^2 J/m^3) = (0.04 kg) * a

a = (-295 x J/kgm^2, 273.75 y^2 J/kgm^3)

As x = 0.40 m and y = 0.60 m,

a = ((-295 J/kgm^2)*(0.4 m), ((273.75 J/kgm^3)*(0.6 m)^2)

a = (- 118 m/s^2, 98.55 m/s^2)

A] Magnitude of a, = sqrt(118^2 + 98.55^2)

= 153.74 m/s^2

B] theta = arctan( (98.55 m/s^2) / (-118 m/s^2) ) = - 39.86 degree

x is negative, and y is positive, so theta must be in the second quadrant.

Your calculator however gives an angle in the fourth quadrant.

Now, remember that

tan theta = tan( theta +180)

so to get the angle in the second quadrant, add 180°

theta = -39.86 degree + 180 degree = 140.14 degree   

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