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2 years ago

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A 2 kg ice cube is released from point A and slides on a frictionless track as shown in the figure. Determine the speed of the c

ube at points B and C.

1 answer:

kherson [118]2 years ago

8 0

According to the conservation of energy

Potential energy at any given instance is equal to the Kinetic energy as energy can neither be created nor be destroyed

Mass is 2kg=m

#A

h=5m

PE=mgh

PE

2(9.8)(5)
10(9.8)
98J

Now

KE=98J
1/2mv²=98J
1/2×2v²=98J
v²=98J
v=√98
v=9.4m/s

#B

h=3.2m

PE:-

2(3.2)(9.8)
6.4(9.8)
62.7J

Now

KE=62.7J
1/2mv²=62.7
v²=62.7
v=√62.7
v=7.9m/s

#C

h=2m

PE

2(2)(9.8)
4(9.8)
39.2J

Now

v²=39.2
v=√39.2
v=6.3m/s

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Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

$U = \frac{GMm}{r}$

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

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Point B )

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$U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}$

$U_2= -5.8925*10^9J$

Then,

$\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)$

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