What average power does the boy need to produce?

[chwala]

Homework Statement

see attached Number 3 (highlighted in red)

Relevant Equations

Power and energy

In my approach, i have

$F_{Resolved} = 300 \cos 60^{\circ} = 150$ Newtons.

I also have,

... from $v=u +at$
$2=0 +6a$
$a= \dfrac{1}{3}$.

$F=ma$
$F=30 ×\dfrac{1}{3} = 10$ Newtons

Therefore,

$P= \dfrac{(150+10) ×6}{6} = 160$ W

Insight is welcome. Thanks.

 

[haruspex]

$F_{Resolved} = 300 \cos 60^{\circ} = 150$ Newtons.

What force are you calculating, and where does that angle come from?

I also have,

... from $v=u +at$
$2=0 +6a$
$a= \dfrac{1}{3}$.

We are not told whether the boy starts at rest.

$P= \dfrac{(150+10) ×6}{6} = 160$ N

The Newton is a unit of force, not power.

 

[chwala]

Typo will amend that...

 

[haruspex]

Typo will amend that...

Ok, but my first point in post #2 is the important one.

 

[chwala]

I used this approach,

 

[haruspex]

I used this approach,

View attachment 353813

I ask again, where do you get that angle from?
It is not given in the question, and neither is the distance of 6m.
And you have not answered my question about the force. What forces act on the boy? What is the net force? How does that relate to the acceleration?

I'll give you a hint: forces are not the way to solve the question. What else can you try?

 

[chwala]

I used $s=ut + \dfrac{1}{2}at^2$ to determine the hypotenuse distance... with the vertical height given i was able to determine the angle as indicated. Looks like this was wrong? The forces are weight acting downwards, there is also the normal contact force which acts perpendicularly to direction of motion, therefore no work done and thus will not be considered...and the force acting along the slope which i was able to use $F=ma$ to analyse that.


Ok i see i need to use energy principles... let me check on this again.

 

[chwala]

Now clear we have,

Total energy change = gravitational potential energy + kinetic energy.

$= (30×10×3) + \left[ \dfrac{1}{2} ×30× 4 \right]= 900+60 = 960$ Joules.

Therefore, Power = $\dfrac{960}{6} = 160$ Watts.

Thanks...learnt something today.

 

[haruspex]

I used $s=ut + \dfrac{1}{2}at^2$ to determine the hypotenuse distance... with the vertical height given i was able to determine the angle as indicated.

You are not given s or a, so I do not see how you could have found either.

Now clear we have,

Total energy change = gravitational potential energy + kinetic energy.

$= (30×10×3) + \left[ \dfrac{1}{2} ×30× 4 \right]= 900+60 = 960$ Joules.

Therefore, Power = $\dfrac{960}{6} = 160$ Watts.

Thanks...learnt something today.

Yes, except that, as I noted in post #2, you are not told that this is from a standing start. Maybe he was already doing 2ms^-1 when he reached the bottom of the stairs.

Also you should include units as soon as you introduce numbers:
$(30kg×10ms^{-2}×3m) + \left[ \dfrac{1}{2} ×30kg× 4m^{2} s^{-2} \right]= 900J+60 J= 960J$.

 

[chwala]

... so the ascent is vertical and not an incline- i did not seem to get that right.

 

[phinds]

... so the ascent is vertical and not an incline- i did not seem to get that right.

No, stairs are not vertical. The climb is at an angle but why do you think that matters?

If a man carries 100 lbs over a total vertical distance of 10 feet, is the work done different depending on whether he does this across 10 feet horizontally or 100 feet horizontally?

 

[Lnewqban]

... so the ascent is vertical and not an incline- i did not seem to get that right.

What we know for sure is that the muscular effort of that child has increased the potential energy of 3 kg corresponding to a delta altitude of 3 meters in 6 seconds.

Work does not depend on time, but power does.
If the muscles had needed 60 seconds to perform the same work, we would have said that an average power 10 times lower had been produced instead.

 

[Steve4Physics]

... so the ascent is vertical and not an incline- i did not seem to get that right.

Just to add to what's already been said (since it’s important), note that the change in gravitational potential energy ($\Delta$GPE) is $mg\Delta h$ where $\Delta h$ is the change in height.*

For example, $\Delta$GPE for a mass moving 2 metres vertically upwards is the same as for the mass moving along a ramp where the top is 2 metres higher than the bottom.

The route taken doesn't matter - when finding $\Delta$GPE it's only the height-difference between the start and end points that matters.
___________________
*Assuming the height-change isn’t so big that changes in $g$ need to be taken into account.