Energy: Block sliding down frictionless ramp

[basenne]

1. In Figure 8-49, a block is sent sliding down a frictionless ramp. Its speeds at points A and B are 1.90 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 3.85 m/s. What then is its speed at point B?

Figure 8-49




2. KE = 1/2mv^2

PE = mg (h)



3. 1/2(m)(2.2^2) + m(9.8)(h) = 1/2(m)(2.6^2) + m(9.8)(.5h)

Yes, I know now that I can't assume that the height for point b is .5 of the height for point A. I'm totally lost on this question and I'd really appreciate a push in the right direction.

 

[venkatg]

Note: answer deleted by moderator.

 

[basenne]

Umm, I really appreciate the answer, however, would you care to elaborate how you got it?

And yeah, you're correct. (Although there was never any doubt that you wouldn't be :D)

 

[venkatg]

Note: detailed solution deleted by moderator.

 

[rwisz]

I would approach this problem by first identifying the key variables involved. In this case, we have the mass of the block (m) and the speeds at points A and B (vA and vB). We also have the acceleration due to gravity (g). Using the given equation for kinetic energy (KE = 1/2mv^2), we can set up the following equations for the two scenarios:

Scenario 1: 1/2(m)(1.90^2) + m(9.8)(h) = 1/2(m)(2.60^2) + m(9.8)(h)

Scenario 2: 1/2(m)(3.85^2) + m(9.8)(h) = 1/2(m)(vB^2) + m(9.8)(h)

Since we are trying to solve for the speed at point B, we can rearrange the equation for scenario 2 to isolate vB:

1/2(m)(vB^2) = 1/2(m)(3.85^2) + m(9.8)(h) - m(9.8)(h)

vB^2 = 3.85^2 + 2(9.8)(h)

Now, we need to find the value of h for both scenarios. We can do this by setting the equations for potential energy (PE = mgh) equal to each other and solving for h:

1/2(m)(vA^2) + m(9.8)(h) = 1/2(m)(3.85^2) + m(9.8)(h)

1/2(m)(vA^2) = 1/2(m)(3.85^2)

h = (vA^2 - 3.85^2)/(2(9.8))

Using the given values for vA, we can calculate h for both scenarios. Plugging these values into the equation for vB, we get:

Scenario 1: vB^2 = 3.85^2 + 2(9.8)(0.227) = 4.22^2

vB = 4.22 m/s

Scenario 2: vB^2 = 3.85^2 + 2(9.8)(0.454) =