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A 392-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800MR2. Friction does work on the wheel as it rolls up the hill to a stop, a height h above the bottom of the hill; this work has absolute value 2450 J .


Calculate h.


Express your answer with the appropriate units.

Answer :

boffeemadrid

Answer:

14.392 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

M = Mass = [tex]\dfrac{392}{9.81}[/tex]

[tex]\omega_{0}[/tex] = Angular velocity = 25 rad/s

[tex]v_{0}[/tex] = Initial velocity

[tex]W_f[/tex] = Friction =2450 J

Here the energy is balanced

[tex]\frac{1}{2} I \omega_{0}^{2}+\frac{1}{2} M v_{0}^{2}-W_{f}=M g h[/tex]

[tex]\Rightarrow (0.8)\left(\frac{1}{2}\right) M R^{2} \omega_{0}^{2}+\frac{1}{2} M \omega_{0}^{2} R^{2}-W_{f}=M g h[/tex]

[tex]\Rightarrow h=\dfrac{\frac{1}{2} I \omega_{0}^{2}+\frac{1}{2} M v_{0}^{2}-W_{f}}{Mg}\\\Rightarrow h=\frac{(0.8)\left(\frac{1}{2}\right)\left(\frac{392}{9.81}\right)(0.6)^{2}(25)^{2}+\left(\frac{1}{2}\right)\left(\frac{392}{9.81}\right)(25)^{2}(0.6)^{2}-2450}{392}\\\Rightarrow h=14.392\ m[/tex]

The h value is 14.392 m

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