A force in the xy plane is given by: F= - (b/r(adsbygoogle = window.adsbygoogle || []).push({}); ^{3}) (x i^{^}+ y j^{^}) where b is a positive constant and r= sqrt(x^{2}+ y^{2})

a) Show that the maginitude of the force varies as the inverse of the square of the distance to the origin and that its direction is antiparallel (opposite) to the radius vector r= x i^{^}+ y j^{^}).

Could anybody please verify my work?:

The following needs to be true:

F ~ 1/r^{2}=> F= c /r^{2}, where c is a constant.

F * F= F^{2}= (b^{2}x^{2}/r^{6}) + (b^{2}y^{2}/r^{6})

F= sqrt((b^{2}x^{2}/r^{6}) + (b^{2}y^{2}/r^{6}) )= sqrt((b^{2}/r^{6}) (x^{2}+ y^{2})) = b/r^{3}sqrt(x^{2}+ y^{2})

Because r= sqrt(x^{2}+ y^{2}) substitution gives:

b/(sqrt(x^{2}+ y^{2}))^{3}sqrt(x^{2}+ y^{2})= b/ (x^{2}+ y^{2}).

r^{2}= (x^{2}+ y^{2}) so F= c /r^{2}is true where the constant c= the constant b in this case.

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F= - (b/r^{3}) (x i^{^}+ y j^{^}) needs to be antiparallel to r= x i^{^}+ y j^{^}).

If two vectors are parallel the following needs to be true: AB (dotproduct)= AB (length vectors).

The dotproduct AB gives: -bx^{2}/r^{3}+ -by^{2}/r^{3}

Substitution of r= sqrt(x^{2}+ y^{2}) and adding the two terms gives:

(-bx^{2}-by^{2})/(sqrt(x^{2}+ y^{2}))^{3}= (-b (x^{2}+ y^{2}))/(sqrt(x^{2}+ y^{2}))^{3}= -b/(x^{2}+ y^{2})

The length of F= b/(x^{2}+ y^{2}), calculated in the first part of the question.

The length of r= r * r= r^{2}= x^{2}+ y^{2}

r= sqrt(x^{2}+ y^{2})

Fr (length of the vectors) gives: b (sqrt(x^{2}+ y^{2})/(x^{2}+ y^{2}) = b/ (sqrt(x^{2}+ y^{2}))

Therefore it is true that Fr (dotproduct)= Fr (length vectors):

-b/(x^{2}+ y^{2})= b/ (sqrt(x^{2}+ y^{2})) with the only restriction that Fr (dotproduct) is negative and therefore F and r are antiparallel.

Ok till now I get the problem, but I'm not sure how to solve b, c and d... so please help me ...

b) If b= 3 Nm^{2}find the work done by this force on a particle moving along a straight line path between an initial position x= 2 m, y= 0 m and a final position x= 5 m, y= 0 m.

I know W= Fs and s= 3 m but I don't know how to calculate the force with - (3/r^{3}) (x i^{^}+ y j^{^})...

c) Find the work done by this force on a particle moving once around a circle of radius r= 7 mcentered around the origin.

d) If this force is the only force acting on the particle what's the particle's speed as it moves along this circular path? Assume that the particle's mass is m= 2 kg.

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# Homework Help: How to: Force, work & dot product

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