Answer:
[tex]v = \sqrt{v_{o}^{2}+2\cdot g\cdot a\cdot (x_{o}^{2}-x^{2})}[/tex],[tex]N = m\cdot \left[\frac{v_{o}^{2}+2\cdot g \cdot a \cdot (x_{o}^{2}-x^{2})}{\frac{\left[1 + 4\cdot a^{2}\cdot x^{2} \right]^{\frac{3}{2} }}{2\cdot |a|} } + \frac{g}{\sqrt{1+4\cdot a^{2}\cdot x^{2}}} \right][/tex], [tex]a = g \cdot \frac{2\cdot a \cdot x}{\sqrt{1+4\cdot a^{2}\cdot x^{2}}}[/tex]
Explanation:
An expression for the velocity can be derived by applying the Principle of Energy Conservation:
[tex]\frac{1}{2}\cdot m \cdot v_{o}^{2} + m\cdot g \cdot y_{o} = \frac{1}{2}\cdot m \cdot v^{2} + m\cdot g \cdot y[/tex]
[tex]\frac{1}{2} \cdot v_{o}^{2} + g \cdot y_{o} = \frac{1}{2} \cdot v^{2} + g \cdot y[/tex]
[tex]v = \sqrt{v_{o}^{2}+2\cdot g\cdot (y_{o}-y)}[/tex]
Let assume that chute is frictionless, so that box is moved because of gravity. The equations of equilibrium for the box are:
[tex]\Sigma F_{x'} = m\cdot g \cdot \sin \theta = m \cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = m\cdot \frac{v^{2}}{\rho}[/tex]
The expression for the radius of curvature and trigonometric functions are, respectively:
[tex]\sin \theta = \frac{dy}{ds} = \frac{1}{\sqrt{1+(\frac{dy}{dx} )^{2}} }\cdot \frac{dy}{dx}[/tex]
[tex]\cos \theta = \frac{dx}{ds} = \frac{1}{\sqrt{1+(\frac{dy}{dx} )^{2}}}[/tex]
[tex]\rho = \frac{\left[1 + (\frac{dy}{dx} )^{2} \right]^{\frac{3}{2}} }{\left|\frac{d^{2}y}{dx^{2}}\right| }[/tex]
Let assume that chute has the following form:
[tex]y = a\cdot x^{2}[/tex]
First and second derivatives of the function are, respectively:
[tex]\frac{dy}{dx}= 2\cdot a \cdot x[/tex]
[tex]\frac{d^{2}y}{dx^{2}} = 2\cdot a[/tex]
Trigonometric functions and radius of curvature are:
[tex]\sin \theta = \frac{2\cdot a \cdot x}{\sqrt{1+4\cdot a^{2}\cdot x^{2}} }[/tex]
[tex]\cos \theta = \frac{1}{\sqrt{1+4\cdot a^{2}\cdot x^{2}} }[/tex]
[tex]\rho = \frac{\left[1+4\cdot a^{2}\cdot x^{2}\right]^{\frac{3}{2} }}{2\cdot |a|}[/tex]
The velocity as a function of x is:
[tex]v = \sqrt{v_{o}^{2}+2\cdot g\cdot a\cdot (x_{o}^{2}-x^{2})}[/tex]
The normal force on the box as a function of x is:
[tex]N = m \cdot \left[ \frac{v^{2}}{\rho} + g\cdot \cos \theta \right][/tex]
[tex]N = m\cdot \left[\frac{v_{o}^{2}+2\cdot g \cdot a \cdot (x_{o}^{2}-x^{2})}{\frac{\left[1 + 4\cdot a^{2}\cdot x^{2} \right]^{\frac{3}{2} }}{2\cdot |a|} } + \frac{g}{\sqrt{1+4\cdot a^{2}\cdot x^{2}}} \right][/tex]
The tangential acceleration on the box as a function of x is:
[tex]a = g\cdot \sin \theta[/tex]
[tex]a = g \cdot \frac{2\cdot a \cdot x}{\sqrt{1+4\cdot a^{2}\cdot x^{2}}}[/tex]
