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andrew-mc [135]
3 years ago
6

A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t

he work-kinetic energy theorem and disregarding friction, find the
velocity of the crate at the bottom of the ramp. (g = 9.81 m/s?)
Physics
1 answer:
Elis [28]3 years ago
3 0

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

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Answer:

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(a)

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ωf = I' ω' /  ( I' + 5I' )

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(b)

Initial kinetic energy will be K'

K' = I'ω'² / 2

and final K.E. will be   K'' =  ( I' + I'' )ωf² / 2

                                   K'' = ( I' + 5I' ) (ω'/ 6)²/ 2

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                                   K'' = I' ω'²/ 12

therefore the fraction lost is

                ΔK/K' = ( K' - K'' ) / K'

                           =  {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)

                            = 5/6

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