A 2.13-kg solid sphere (radius 0.135 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.703 m high and 5.07 m long. When the sphere reaches the bottom of the ramp, whatis its total kinetic energy?=Tries 0/6Submit AnswerWhat is its rotational kinetic energy?Submit AnswerTries 0/6What is its translational kinetic energy?

Question
Asked Oct 29, 2019
A 2.13-kg solid sphere (radius 0.135 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.703 m high and 5.07 m long. When the sphere reaches the bottom of the ramp, what
is its total kinetic energy?
=
Tries 0/6
Submit Answer
What is its rotational kinetic energy?
Submit Answer
Tries 0/6
What is its translational kinetic energy?
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A 2.13-kg solid sphere (radius 0.135 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.703 m high and 5.07 m long. When the sphere reaches the bottom of the ramp, what is its total kinetic energy? = Tries 0/6 Submit Answer What is its rotational kinetic energy? Submit Answer Tries 0/6 What is its translational kinetic energy?

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Expert Answer

Step 1

At the top of the incline plane, the sphere have only the potential energy, then the total kinetic energy is

Ermgh
total
(2.13kg) (9.8m/s)(0.703m)
=14.67 J
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Ermgh total (2.13kg) (9.8m/s)(0.703m) =14.67 J

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Step 2

As the sphere comes down its potential is converted into kinetic energy (rotational+translational).

- mv2 I@
mgh
v = r@; w
12
1
-mv
2
25
mr2
solid sphere
5
I.
7
gh
10
10
1
7
gh
10
(9.8m/s2)(0.703m)
7
3.13m/s
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- mv2 I@ mgh v = r@; w 12 1 -mv 2 25 mr2 solid sphere 5 I. 7 gh 10 10 1 7 gh 10 (9.8m/s2)(0.703m) 7 3.13m/s

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Step 3

Then the rotational ki...

I0
К.
rot
2
12
т
mr
25
5
1
(2.13kg)(3.13m/s)
5
= 4.17J
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I0 К. rot 2 12 т mr 25 5 1 (2.13kg)(3.13m/s) 5 = 4.17J

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