# Inclined Plane Homework: Calculating Speeds at Photogates 1 and 2

• anton717
In summary: Okay, so let's recap what we have done here: we have taken the given information and applied the principle of conservation of energy to find the velocity at photogate 1. We have also discussed using the equation vf^2=vi^2+2ad and the fact that acceleration is constant on an inclined plane. We have also talked about the importance of carefully checking our calculations for mistakes. In summary, we used the given information to apply the principle of conservation of energy and correctly found the velocity at photogate 1. We also discussed the option of using the equation vf^2=vi^2+2ad and the fact that acceleration is constant on an inclined plane. Lastly, we emphasized the importance of checking calculations
anton717

## Homework Statement

Photogate 1 is at the 25.2 cm mark, and photogate 2 is at the 66.4 cm mark along the track, Assume
* The bottom of the track is at U = 0 J (potential energy is zero).
* The total mass of the cart is 214 g.
* The track is totally frictionless

Suppose the cart is released at rest at some point above photogate 2 (the higher point on the track). If the cart passes through photogate 2 at speed 0.497 m/s, at what speed will the cart pass through photogate 1?
v1 = ____________ m/s

## Homework Equations

I am thinking about mgh+ 1/2 (mv^2) = mgh' + 1/2 (mv'^2)

## The Attempt at a Solution

I went ahead and did PEi+KEi=PEf+KEf
and got a wrong answer for Vf (which is v1)? ( I got something like vf= 0.998 )

What have I done wrong? Did I not consider the acceleration?
Well, I tried that way a=gsin(x), I found the a to be 1.143 m/s^2 and after that I am a bit confused with 2 different heights and lack of info to find Vf (v1)... (may be I m not getting it)
Would be awesome if someone would help me with the concept or answer.
Thanks,

What did you use for h when trying to apply conservation of energy?

Also, your statement is missing information in the inclination of the plane.

anton717 said:
that way a=gsin(x)
I don't see you mentioning ##x## anywhere before,

anton717 said:
I am thinking about mgh+ 1/2 (mv^2) = mgh' + 1/2 (mv'^2)
I doubt, you understand what ##h## and ##h'## actually mean. If you just substituted the value of the mark at which the photogates are present, you are wrong, h should be vertical to the ground, seeing the title of this thread i doubt if it is the vertical height that you have used

Orodruin said:
What did you use for h when trying to apply conservation of energy?

Also, your statement is missing information in the inclination of the plane.

Sorry :/, I forgot to write this : " Suppose the inclined plane is tipped at angle θ= 6.7 o. The 0 cm mark is at the bottom. "

Suraj M said:
I don't see you mentioning ##x## anywhere before,

It's Theta but I don't have the symbol so I just wrote x but then I realized that I did not even write the what x is equal to. Sorry. x = 6.7* so it's a=gsin(6.7). My fault.

Suraj M said:
I doubt, you understand what ##h## and ##h'## actually mean. If you just substituted the value of the mark at which the photogates are present, you are wrong, h should be vertical to the ground, seeing the title of this thread i doubt if it is the vertical height that you have used

Okay. Read the question carefully. It gives you 2 heights. But never gives you the actual height of the inclined plane. If you want to find the actual height then you need the total distance of the hypotenuse (at least) but its not given either. Those H' and H are taken from H=dsin(x) and H'=d'(sin(x)) are the heights where each photogate is located. I wrote this because I thought finding P.E and K.E energy at the point where each photogate is located would help me. But the answer I got was wrong...

anton717 said:
mgh+ 1/2 (mv^2) = mgh' + 1/2 (mv'^2)
This equation is correct, you should get the right answer from this equation. What should the answer be? Are you confident about your calculations?

anton717 said:
gives you the actual height of the inclined plane
Which you don't need to know as they have already given you the velocity attained.
anton717 said:
found the a to be 1.143 m/s^2 and after that I am a bit confused
You can do it this way too.
You don't need to be confused, you can assume you're new x-axis parallel to the inclined plane, then you're solution reduces to
anton717 said:

Suraj M said:
This equation is correct, you should get the right answer from this equation. What should the answer be? Are you confident about your calculations?

I said it somewhere in the main post that I got about 0.998 ( got it yesterday, don't really remember the exact #) but it says its wrong ( its online and 1 try is left :p).
Yes I am pretty sure I calculated it correctly. I found each height, (converted it to m and kg) so I don't see the mistake in conversion too.

So according to you the change in PE + change in KE =0 formula is correct right? (formula that is mentioned above)
Then what's wrong...

The thing why I am not sure about vf^2=vi^2 +2ad is that the actual initial v is 0 since it leaves the inc.plane from rest. AND the "a" found is acceleration for the total distance of the inc.plane. It's not "a" from Photogate 2 to Photogate 1 . If you know what I mean. If we assume so the v final (photogate 1) will be much higher...which I think is wrong...

anton717 said:
So according to you the change in PE + change in KE =0 formula is correct right? (formula that is mentioned above)
Then what's wrong...
Please write out your full computation with all numbers. If you do not, then we are just stabbing in the dark as to where you might have gone wrong.

Guys, forget it. I got the right answer. I made a mistake on one little thing.
Absolutely gutted.

anton717 said:
I made a mistake on one little thing. .
And what was that?

Suraj M said:
And what was that?
basically the equation dKE + dPE = 0 (d as delta , change) is correct, but when I was plugging in number for given V (at photogate 2) I plugged in its mass by mistake ...
Its embarrassing lol

## 1. What is an inclined plane?

An inclined plane is a simple machine that consists of a flat surface that is raised at an angle. It is used to reduce the effort needed to lift heavy objects by spreading the force required over a longer distance.

## 2. How do photogates work?

Photogates are sensors that use light beams to detect the passage of an object. They consist of two units, one that emits the light beam and one that receives it. When an object passes through the beam, it breaks the connection and triggers a timer to record the time of passage.

## 3. How do I calculate speed using photogates?

To calculate speed using photogates, you will need to measure the distance between the two photogates and the time it takes for an object to pass through them. Then, you can use the formula speed = distance/time to calculate the speed of the object.

## 4. What is the significance of calculating speeds at photogates 1 and 2?

Calculating speeds at photogates 1 and 2 allows you to determine the acceleration of an object as it moves down an inclined plane. This can help you understand the relationship between the angle of the inclined plane and the acceleration of the object.

## 5. How can I use the data from calculating speeds at photogates 1 and 2 in real-life situations?

The data from calculating speeds at photogates 1 and 2 can be applied in various real-life situations, such as understanding the speed and acceleration of an object moving down a ramp or measuring the effectiveness of different inclined planes in reducing the effort needed to lift heavy objects. It can also be used in engineering and physics experiments to study the effects of gravity and friction on moving objects.

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