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A block of $\rm55g$ is sliding down a ramp of $35^o$ of inclination.

The hypotenuse of the ramp is $\rm63cm$ and the height is $\rm36cm$. $v_i=0$ as the block starts at rest.

I did 3 trials of letting the block slide down the ramp and the time intervals I got each are:

1) $\rm0.41 s$

2) $\rm0.44 s$

3) $\rm0.47 s$

So then I used the $d = v_i\times t + \frac{at^2}{2}$ formula to calculate the acceleration of the block and I got

1) $\rm7.5 m/s^2$

2) $\rm6.5 m/s^2$

3) $\rm5.7 m/s^2$

Then, I used the $v_f = v_i + at$ to find the velocity at the bottom of the ramp.

1) $\rm3.1 m/s$

2) $\rm2.9 m/s$

3) $\rm2.7 m/s$

And then I found the total energy at the top of the ramp , which would only be the potential energy as initial velocity is zero. So it's $\rm0.19J$ that I calculated.

Then when I move to solve the total energy at the bottom of the ramp, there is a problem. Potential energy is zero and there's only kinetic energy, and also final energy should be smaller than initial energy because of friction, but I keep getting a greater value for all of them, as well as my change in mechanical energy, which should be negative and I keep getting a positive value. Please help.

initial energy $\rm0.19 J$

final energy (3 trials) calculated by $mv^2/2$

1) $\rm0.26 J$ (it's greater that EI but should not be!)

2) $\rm0.23 J$

3) $\rm0.2 J$

*my teacher said that I should be getting a negative value for change of energy and she hasn't taught us how to do include experimental errors..she said the errors shouldn't affect the results like that so there must be something wrong with my process, but I can't figure out what it is?

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  • $\begingroup$ Are you using g as a? You are on an incline so the acceleration is not just g! $\endgroup$ – yankeefan11 Jul 23 '15 at 20:54
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    $\begingroup$ You should be taking into account the errors introduced in measuring the initial height of the object (including estimating the error from not measuring exactly at the center of mass), the distance down the ramp, the angle of the ramp, the travel time, and the mass. That's not to say that you didn't make some other mistake in your calculation, but there's no way to reasonably compare the prediction to the experiment without keeping track of the experimental errors. $\endgroup$ – d_b Jul 23 '15 at 20:56
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    $\begingroup$ I only used g when I calculated the initial energy(potential energy) which would be mgh so 0.055g*9.8m/s^2*0.36m. Is that right? $\endgroup$ – stacope32 Jul 23 '15 at 20:57
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    $\begingroup$ Yes that is your initial energy $\endgroup$ – yankeefan11 Jul 23 '15 at 21:04
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    $\begingroup$ Quick tip: in the frictionless case, the acceleration down a slope is $a=g\sin\theta$ where $\theta$ is the angle of the slope to the horizontal. Using the figures you've provided for your setup, this works out to be 5.6 m/s$^2$. Whatever you calculate your acceleration to be, it cannot be larger than this value. So that strongly suggests that it's your measurement of the system, or more likely your timing, which is in error. $\endgroup$ – tok3rat0r Jul 24 '15 at 13:34
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tok3rat0r probably has the right of it.

You have not told us exactly how your timing data was acquired. If it was done by some sort of stopwatch (mechanical or electrical) you should assume an uncertainty of at least 0.1 seconds, and perhaps more for a situation where you have to push a button twice in 1/2 second.

If you assume a worst-case error of 0.1 seconds, your timing measurement becomes 0.54 seconds, and if you redo your calculations you'll get a final energy of 0.15 J. While this does not conclusively prove that your timing technique was the problem, it suggests that you should look very closely at it.

One approach would be to increase your ramp height so that your duration gets much longer - several seconds at a minimum. If you are using an electronic timer, you could replace your timing switch with two switches - one at the top of the ramp and one at the bottom, with a mechanical block release such as burning through a thread.

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As user37496 said, the proper way is to calculate the error, using the errors of your measurements. But there is also an alternative, that is you measure often and calculate the mean, and determine the standard deviation as an error. Three times is not really enough to give a reasonable result but it shall do.

The mean time is: $t = \frac{1}{3}(0.41 + 0.44 + 0.47)s \approx 0.44s$

Now I don't know if you need the intermediate results but the proper way to calculate the energy is to first derive the formula and the put the measured values in. We have $ d = \frac{1}{2}at^2 \Rightarrow a = \frac{2d}{t^2}$ and $v= at$ thus $v = \frac{2d}{t^2}t = \frac{2d}{t}$, finally with kinetic energy $T = \frac{1}{2}mv^2$

$$T = \frac{1}{2}m\big(\frac{2d}{t}\big)^2 = 2m \big(\frac{d}{t}\big)^2$$

Inserting the values: $ T \approx 0.23J$

The initial potential energy is $ V = mgh \approx 0.19J$

Thus the energy difference is $\Delta E = T -V = 2m \big(\frac{d}{t}\big)^2 - mgh \approx 0.03J$

This result is in contradiction to the expected negative energy difference, you can only conclude that either the errors in your measurement cause the false result, or that the model (constant acceleration due to constant drag) is not applicable, or both. My money is on the first explanation.

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There doesn't seem to be anything wrong with your calculations given the data and assumptions you've made. So lets consider how long you would expect the block to take in a frictionless situation.

By conservation of energy, $mgh=\frac{1}{2}mv_f^2$.

Putting in your values I get $v_f=2.63 ms^{-1}$.

Then using $d=\frac{1}{2}(v_i+v_f)t$

I get an expected time of $t=0.48s$.

Depending on your set-up, if you are measuring time with a stop-watch by eye I think 0.1s is a pretty reasonable uncertainty in your time measurements. You could then claim your difference is zero within uncertainty.

In reality friction should also play a role and result in a loss of energy and therefore longer times. It would appear that for your set-up the effect of friction is pretty small. You could 'fix' your experiment by using a higher friction surface and/or shallower slope so that the effect of friction is increased and overwhelms any other influences due to uncertainty, etc.

Another issue may be that you give the block some initial speed unintentionally when releasing it. This can be difficult to account for and depends on how the experiment is set-up.

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You must also take into account the dimensions of the block. You have to be careful about how much distance the block actually covers. If the block starts from rest as shown here:enter image description here

Then the block actually covers $(63-x)\rm cm$ where $x$ is the length or breath of the block whichever is parallel to the motion. As I can see from your calculations that you took the distance to be $63\rm cm$. The acceleration should be lesser than you calculated if you subtract that $x$ and so be the final velocity $v_f$.

See the time very carefully with a more precise clock and repeat many times.

If the inclination is $35^o$ then the component of acceleration due to gravity in the direction of motion is $g\rm cos55^o=5.7ms^{-1}$. So, your last observation is much closer.

Lastly, since the block still remains at some height from the base of the ramp, you should take the difference of the initial and final potential energy of the block because that much potential energy is converted to kinetic energy. Kinetic energy should come out to be a little lesser since some energy will be dissipated as heat due to friction, drag etc.

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protected by Qmechanic Sep 7 '15 at 5:37

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