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A 6.71 kg object has the following acceleration components:

ax = ( 0.51 m/s2) + ( 0.75 m/s3) t

ay = ( 11.9 m/s2) + ( 0.59 m/s3) t

A) What is the magnitude of the net force acting on the object at time t = 6.07 s? In Newtons: ______

B) What is the direction of the net force at this same time? Give your answer as a number of degrees counter-clockwise from the x axis. In Degrees: _________

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Added Fri, 18 Dec '15
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You have to add vectors, taking the angle between them into account.

At t= 6.07 s

a_x = ( 0.51 m/s2) + ( 0.75 m/s3) *6.07= 5.0625 m/s�

a_y = ( 11.9 m/s2) + ( 0.59 m/s3) *6.07= 15.48 m/s�

You can add these components and use F=ma, where 'a' is the resultant acceleration. or you can work out Fx and Fy and combine these. It doesn't' make any difference....

F_x = m*a_x = 6.71 x 5.0625 = 33.96N

F_y = m*a_y = 6.71 x 15.48= 103.87 N

|F| = ?(F_x� + F_y�)

= ?(33.96� + 103.87�)

= 109.28N

Angle = tan-1(F_y/F_x) = tan-1(103.87/33.98) = 71.88?

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Added Fri, 18 Dec '15
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