Exercise 7.37 A small block with mass 0.0400kg is moving in the xy -plane. The n

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Exercise 7.37

A small block with mass 0.0400kg is moving in the xy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.55J/m2 )x2-(3.55J/m3 )y3.

Part A

What is the magnitude of the acceleration of the block when it is at the point x= 0.20m , y= 0.59m ?

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= 0.20m , y= 0.59m ?

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Exercise 7.37

A small block with mass 0.0400kg is moving in the xy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.55J/m2 )x2-(3.55J/m3 )y3.

Part A

What is the magnitude of the acceleration of the block when it is at the point x= 0.20m , y= 0.59m ?

Express your answer with the appropriate units.

a =

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Part B

What is the direction of the acceleration of the block when it is at the point x= 0.20m , y= 0.59m ?

? =   ? counterclockwise from the +x-axis

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Explanation / Answer

First determine the force vector. This vector is ( - gradient of U)

F = ( - dU/dx , - dU/dy )
= ( -5.55 *2 x, 3.55*3*y^2) N
= ( -11.2 x, 10.65 y^2) N

a = F/m,

m= 0.0400 kg

a = ( - 280x, 266.25 y^2) m/s^2

(A) substituting x=0.20 and y=0.59 gives

a = (-56 , 157.08) m/s^2

(B) the angle follows from

tan(angle) = a_y / a_x = 157.08/(-56) = -1.7257
angle = -70.38 degrees

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