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Confirm my logic on power

  • Thread starter Winner
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  • #1
94
8
Ok,

So, lets 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:
 

Answers and Replies

  • #2
203
0
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
94
8
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
31
0
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
22
0
perfect!!! so the force is 40 * 0.4...and that by 5 gives you the power!
 
  • #6
94
8
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.
 

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