A wheel with a weight of 396 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 22.7 rad/s. The radius of the wheel is 0.569 m and its moment of inertia about its rotation axis is 0.800 MR^2 . Friction does work on the wheel as it rolls up the hill to a stop, at a height of above the bottom of the hill; this work has a magnitude of 3476 J . Calculate "h" Use 9.81 for the acceleration due to gravity.

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-08-23T09:03:54+02:00

Answer:

The height is 5.67 m.

Explanation:

Given that,

Weight = 396 N

Angular velocity = 22.7 rad/s

Radius = 0.569 m

Moment of inertia = 0.800 kg-m²

Work = 3476 J

We need to calculate the kinetic energy

Using formula of kinetic energy

[tex]K.E =\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]

Put the value into the formula

[tex]K.E=\dfrac{1}{2}\times\dfrac{369}{9.81}\times0.569\times(22.7)^2+\dfrac{1}{2}\times0.800\times(22.7)^2[/tex]

[tex]K.E=5720.43\ J[/tex]

We need to calculate the renaming energy

[tex]E=5720.43-3476=2244.43\ J[/tex]

We need to calculate the height

Using formula of potential energy

[tex]mgh =2244.43[/tex]

Put the value into the formula

[tex]h=\dfrac{2244.43}{\dfrac{396}{9.81}\times9.81}[/tex]

[tex]h=5.67\ m[/tex]

Hence, The height is 5.67 m.