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andreyandreev [35.5K]
3 years ago
9

A solid ball, a solid disk, and a hoop, all with the same mass and the same radius, are set rolling without slipping up an incli

ne, all with the same initial linear (translational) speed. Which goes farthest up the incline?
a)They all roll to the same height
b)The hoop and the disk roll to the same height, farther thank the ball
c)The disk
d)The hoop
e)The ball
Physics
1 answer:
MAXImum [283]3 years ago
7 0

Answer:

d ) is the answer.

Explanation:

Let M be the mass  and R be the radius of each of ball , hoop and disc.

kinetic energy of sphere - 1/2 MV² + 1/2 I ω² ,ω is angular velocity and

V = ωR

kinetic energy of sphere - 1/2 MV² + 1/2 x 2/5 MR² ω²

= 1/2 MV² + 1/5 MR² ω²

MV² ( 1/2 + 1/5 )

= .7 MV²

kinetic energy of Disk - 1/2 MV² + 1/2 I ω² ,ω is angular velocity and

V = ωR

kinetic energy of Disk - 1/2 MV² + 1/2 x 1/2 MR² ω²

= 1/2 MV² + 1/4 MR² ω²

MV² ( 1/2 + 1/4 )

= .75 MV²

kinetic energy of Hoop - 1/2 MV² + 1/2 I ω² ,ω is angular velocity and

V = ωR

kinetic energy of hoop - 1/2 MV² + 1/2  MR² ω²

= 1/2 MV² + 1/2 MR² ω²

MV² ( 1/2 + 1/2 )

=  MV²

Kinetic energy is largest in case of hoop and least in case of sphere . So hoop will go up to the highest point and sphere will go to a height which will be least among the three.

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If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and t
STatiana [176]

Answer:

The velocity is  4.6 m/s^2

Explanation:

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Force = 500N

Distance  s= 0

To find :

Its velocity at s = 0.5 m

Solution:

\sum F_{x}=m a

F\left(\frac{4}{5}\right)-F_{S}=13 a

500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a

400-(500 s)=13 a

a = \frac{400 -(500s)}{13}

a = (30.77 -38.46s) m/s^2

Using the relation,

a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}

a=v \frac{d v}{d s}

v d v=a d s

Now integrating on both sides

\int_{0}^{v} v d v=\int_{0}^{0.5} a d s

\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s

\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}

\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]

\left[\frac{v^{2}}{2}\right]=[15.385-4.807]

\left[\frac{v^{2}}{2}\right]=10.578

v^{2}=10.578 \times 2

v^{2}=21.15

v = \sqrt{21.15}

v = 4.6 m/s^2

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3 years ago
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