A small block with mass 0.0400 kg is moving in the xy -plane. The net force on t

Question

A small block with mass 0.0400 kg is moving in thexy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.95 J/m2 )x2-(3.45 J/m3 )y3.

Part A.

What is the magnitude of the acceleration of the block when it is at the point x= 0.23 m , y= 0.58 m ?

Express your answer with the appropriate units.

Part B.

What is the direction of the acceleration of the block when it is at the point x= 0.23 m , y= 0.58 m ?

Hyun Jin NUpper D vision writing Pro MasteringPhysics: Course MasteringPhysics: Ch 07| ewhile A Roofer ls working eAForce Parallel To The X-x x x x x -> S https://session.masteringphysics.com/myct/itemview?offset-next&assignmentProblemlD;=59703848 :: Apps Thanks for using C Food in Thailand-T -scholar The secret to Sams a Amazon.com: Doubl a Amazon.com: Doubl Physics 220A, Dr. Taheri, Spring 2016 Ch 07 HWExercise 7.33 a Amazon.com: Doubl a Amazon.com: Allegr Other Bookmarks Signed in as Hyun Jin Oh Help Close Resources « previous | 21 of 25 | next Exercise 7.33 Part A 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(z, y) = (5.95 J/m2 )z2-(3.45 J/m"W3 What is the magnitude of the acceleration of the block when it is at the point z = 0.23 m , y = 0.58 m ? Express your answer with the appropriate units aValue Units Submit My Answers Give Up Part B What is the direction of the acceleration of the block when it is at the point x = 0.23 m , y 0.58 m ? counterclockwise from the +x-axis Submit My Answers Give Up ovide Feedba Continue

Explanation / Answer

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

F = ( - U/x , - U/y )
= ( -5.95 *2 x, 3.45*3*y^2) N
= ( -11.9 x, 10.35 y^2) N
where I already have taken out the units, so x and y are numbers.
Because a = F/m, a division by m= 0.0400 kg yields the acceleration vector

a = ( - 297.5 x, 258.75 y^2) m/s^2

(A) substituting x=0.23 and y=0.58 gives

a = (-68.425, 87.04) m/s^2

(B) the angle follows from

tan(angle) = a_y / a_x = -1.27
angle = -51.78 degrees=120.09 degrees

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