Question

When the 5 -kg box reaches point A it has a speed vA=10v_{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

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The equation of the curve is

x12+y12=3x^{\frac{1}{2}}+y^{\frac{1}{2}}=3

dydx=2(3x12)(12x12)\therefore \dfrac{dy}{dx}=2(3-x^{\frac{1}{2}})\left(-\dfrac{1}{2}x^{-\frac{1}{2}}\right)

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