Confirm my logic on power

In summary, Ok, so based on Xiankai's calculations, if you have a coefficient of friction then you would use (40kg x 0.4 m/s^2) - u (mu) x F normal (m x g).
  • #1
94
8
Ok,

So, let's say I was to drag a 40kg load in a set distance of 10 meter. I record the time, it was 5 secs.

Now I find power, P=work/t. We know work=force x distance.

SO my question is, is force just mass of load x acceleration due to gravity? 9.81m/s^2.

So P= ((40kg x 9.81) * 10 meters)/5 seconds,

so 784 watts or so. Thanks guys. What about force of friction? I guess this question doesn't want me to go that far in yet.

Thanks guys. :smile:
 
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  • #2
I don't quite agree with the way you get work and force.
The work you done is against the friction force, not the weight of the load.
Hence force = mg is not applicable.
In fact, work needed to drag an object for different surface is different due to difference in coefficient of friction.
Greater mass do need greater force to drag, but that is because the friction is higher, as the frictional force is greater for a heavier object.
 
  • #3
Harmony said:
I don't quite agree with the way you get work and force.
The work you done is against the friction force, not the weight of the load.
Hence force = mg is not applicable.
In fact, work needed to drag an object for different surface is different due to difference in coefficient of friction.
Greater mass do need greater force to drag, but that is because the friction is higher, as the frictional force is greater for a heavier object.

So how does one do it in terms of the equation?
 
  • #4
a = (v-u)/t

given u = 0, and v = s/t

a = s/t^2

= 0.4 ms^-2

(assuming non-frictional planar flat surface, linear acceleration)

thats what i think, at least
 
Last edited:
  • #5
perfect! so the force is 40 * 0.4...and that by 5 gives you the power!
 
  • #6
I see. So V=d/t, then use that as a final V in a=Vf-Vi/t, a=.4 m/s^2 on a non frictional surface.

Sorry, I just didn't get the formulas Xiankai was using.

I see! so if I had a coefficient of friction then I all I would do is go (40kg x 0.4 m/s^2) - u (mu) x F normal (m x g)? Excellent. Then I'll probably have to consider the static or kinetic friction too... lol, lucky I don't have to do this.
 

1. What is the concept of power in scientific research?

Power in scientific research refers to the ability of a study to detect a true effect or relationship between variables. It is measured by the likelihood of rejecting the null hypothesis when it is false, also known as the study's statistical significance.

2. Why is it important to confirm the power of a study?

Confirming the power of a study is important because it ensures that the research is capable of detecting the effect or relationship being studied. This helps to avoid drawing false conclusions or wasting resources on a study that is underpowered and unlikely to produce meaningful results.

3. How is power calculated in a study?

Power in a study is typically calculated through statistical analysis, using factors such as the sample size, effect size, and alpha level. Power analysis can also be used to determine the necessary sample size for a study to have adequate power.

4. Is a higher power always better in a study?

While a higher power is generally desired in a study, it is not always better. A very high power can result in detecting even small, insignificant effects and can lead to type I errors. It is important to balance power with other factors such as the study's research question and resources available.

5. How can a researcher increase the power of their study?

A researcher can increase the power of their study by increasing the sample size, using more sensitive measures or methods, and reducing sources of error or variability. Additionally, conducting a power analysis before starting the study can help determine the necessary sample size for adequate power.

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