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Question

When the 5 -kg box reaches point A it has a speed $v_{A}=10$ m/s. Determine how high the box reaches up the surface before it comes to a stop. Also, what is the resultant normal force on the surface at this point and the acceleration? Neglect friction and the size of the box.

Solution

VerifiedStep 1

1 of 5The equation of the curve is

$x^{\frac{1}{2}}+y^{\frac{1}{2}}=3$

$\therefore \dfrac{dy}{dx}=2(3-x^{\frac{1}{2}})\left(-\dfrac{1}{2}x^{-\frac{1}{2}}\right)$

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