(adsbygoogle = window.adsbygoogle || []).push({}); Conservation of energy, PE & KE problem. help??

a block is sent sliding down a frictionless ramp. Its speeds at points A and B are 2.00 m/s and 2.60 m/s, respectively. Next, it is again sent sliding down the ramp, but this time its speed at point A is 4.00 m/s. What then is its speed at point B?

I understand the first step is to calculate the KE

KE at pt A:

KEa = (1/2) m(2)^2

KEa= 2m*

calculating the KE at pt B:

KEb = (1/2)mv^2

KEb = 3.38m

This part from the solution guide that I'm confused about:

"the difference between the KE at pt B and at pt A is the gravitational potential energy, therefore

Eg = KEb - KEa

Eg = 3.38 - 2m

Eg = 1.38m (m is the mass)"

I don't understand this part!

So PE = KEa - KEb??

Why is that?

And then (this part I also do not understand)

KEa2 = 1/2m(4.00)^(2) = 8m

To get the answer it would be

8m+1.38m=9.38m

1/2mv^(2) = 9.38m

V= 4.33 m/s

I am very confused. Can someone please explain to one to me. Thanks !

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# Homework Help: Conservation of energy, PE & KE problem. help?

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