Calculating the Work Required to Climb Stairs

[Villager27]

Homework Statement

A 60.78kg student climbs 4.08m of stairs in 3.85s. Calculate the work required to climb the stairs.

m= 60.78kg, Δd_y=4.08m, Δt=3.85s

Homework Equations

W=FΔd OR E_K=1/2mv²

The Attempt at a Solution

My friends and I are conflicted in how to solve this problem. I believe that because it is the work required, that I must use the first equation, and find the force of gravity (the student only needs to overcome gravity to climb the stairs):

W=FΔd
W=(m)(g)(Δd)
W=(60.78)(9.8)(4.08)
W=2430.228J
W=2.43x10³J

My friends, on the other hand, believe that we should use the velocity to find the kinetic energy to climb the staircase:

E_K=1/2mv
E_K=1/2(60.78)(Δd_y/Δt)
E_K=1/2(60.78)(4.08/3.85)
E_K=1/2(60.78)(1.06)
E_K=32.2134J
E_K=32.2J

Which one of us are correct? If neither of us are, could you please provide the solution to this problem, or an explanation on how to get there? Thanks in advance!

 

[PeroK]

Homework Statement

A 60.78kg student climbs 4.08m of stairs in 3.85s. Calculate the work required to climb the stairs.

m= 60.78kg, Δd_y=4.08m, Δt=3.85s

Homework Equations

W=FΔd OR E_K=1/2mv²

The Attempt at a Solution

My friends and I are conflicted in how to solve this problem. I believe that because it is the work required, that I must use the first equation, and find the force of gravity (the student only needs to overcome gravity to climb the stairs):

W=FΔd
W=(m)(g)(Δd)
W=(60.78)(9.8)(4.08)
W=2430.228J
W=2.43x10³J

My friends, on the other hand, believe that we should use the velocity to find the kinetic energy to climb the staircase:

E_K=1/2mv
E_K=1/2(60.78)(Δd_y/Δt)
E_K=1/2(60.78)(4.08/3.85)
E_K=1/2(60.78)(1.06)
E_K=32.2134J
E_K=32J

Which one of us are correct? If neither of us are, could you please provide the solution to this problem, or an explanation on how to get there? Thanks in advance!

It depends whether he still has KE at the top of the stairs. If he has no KE (and has used gravity to slow himself down), then your solution is correct. And, I suspect, this is the answer this is sought.

Alternatively, if he was already moving at a speed $v$ when he reached the stairs and maintained that speed as he climbed and continued at that speed at the top of the stairs, then, again, your answer is correct.

If, however, he has more or less KE at the top of the stairs than he did at the bottom, then that's a different problem. Note that in this case, the speed wouldn't be the vertical speed, but his overall speed, including vertical and horizontal components. This is not, I believe, the intended problem.

 

[TomHart]

Using the second method - i.e., if the amount of energy was only dependent on speed - it would require the same amount of energy to climb 1 stair as it would to climb the stairs of the empire state building.

Edit: P.S. Welcome to Physics Forums.

 

[Villager27]

Thanks for the assistance!

 

[Villager27]

A later part of the question states: Calculate the power the student generated climbing the stairs. Is the following solution correct?

E_M=E_K+E_G
E_M=1/2mv²+E_G
E_M=(1/2)(60.78)(Δd/Δt)²+2430.228
E_M=(1/2)(60.78)(4.08/3.85)²+2430.228
E_M=(1/2)(60.78)(1.06)²+2430.228
E_M=(1/2)(60.78)(1.1236)
E_M=34.146+2430.228
E_M=2464.644J
E_M=2460J

P=E_M/Δt
P=2464.644/3.85
P=640.17W
P=6.40x10²W

Thanks again in advance!

 

[haruspex]

It depends whether he still has KE at the top of the stairs

Yes and no.
If no remnant KE is required then yes, the first method is right and the second wrong. But if we are supposed to assume constant velocity then neither is right; in fact, we do not have enough information. These are stairs, not a ladder, so the velocity will not be purely vertical.

Calculate the power the student generated climbing the stairs. Is the following solution correct?

I would interpret this as meaning average power, so you can use the work done calculated previously and just divide that by time.
If you interpret this as implying the power is constant then the question gets a lot more involved. It would not correspond to constant velocity, nor to constant acceleration, but something between the two.

 

[Villager27]

so you can use the work done calculated previously and just divide that by time.

The first question is asking about the work required, not the actual work performed by the student. I found the mechanical energy so it also includes the additional energy (via kinetic energy) that they've inputted. If I simply divided the work found in the first question by time, I would find the power required, not the actual power put in.

 

[haruspex]

I would find the power required, not the actual power put in.

As I explained, you have no way to determine the average power put in. You have to make other assumptions. If you assume constant speed then you also need to assume a value for the slope.
The first part asks for the work required, implying the minimum value. As PeroK noted, the student can go at whatever speed until close enough to the top that KE alone will take her the rest of the way, arriving with no wasted KE. In the extreme case, the student accelerates at a superhuman rate for a fraction of a second, then makes it the rest of the way expending no further effort. Since that would get the time under one second, it is an allowed solution.
When it asks for the power generated, it is clearly referring to the same context, namely getting to the top of the stairs in the stated time using the least work. To use a larger value for the work than calculated in the first part is simply inconsistent.

 

[PeroK]

The first question is asking about the work required, not the actual work performed by the student. I found the mechanical energy so it also includes the additional energy (via kinetic energy) that they've inputted. If I simply divided the work found in the first question by time, I would find the power required, not the actual power put in.

You asked whether method a) or b) was correct. Three of us explained that method a) was correct, you thanked us for the explanation and then went on to use method b).

 

[Villager27]

went on to use method b)

I used method a) and b) because I thought it was the correct way to figure out the power that the student generated. As by the explanations above, could someone explain why it's only the gravitational potential energy divided by time?

When it asks for the power generated, it is clearly referring to the same context, namely getting to the top of the stairs in the stated time using the least work. To use a larger value for the work than calculated in the first part is simply inconsistent.

Or is this the explanation why?

 

[haruspex]

could someone explain why it's only the gravitational potential energy divided by time?

Because the student can arrange to reach the top of the stairs in the stated time with no residual KE, so all the work has gone into raising the PE.

 

[NascentOxygen]

My friends, on the other hand, believe that we should use the velocity to find the kinetic energy to climb the staircase:

EK=1/2mv
EK=1/2(60.78)(Δdy/Δt)

Is this a correct expression for kinetic energy?

 

[Villager27]

I made a mistake in that expression, but I forgot to fix it. I'll fix it later. @NascentOxygen

 

[TomHart]

I found the mechanical energy so it also includes the additional energy (via kinetic energy) that they've inputted.

I understand your argument for adding in the kinetic energy, as it is definitely energy that the student had to generate at some point. However, I suspect you are making the problem more difficult than it was intended. If I was grading your work, I probably would give you full credit if you worked it that way correctly. But someone else who took a more hard line might argue, "How many people have you seen who stop at the bottom of the stairs before they ascend the stairs? That kinetic energy had nothing to do with work the student generated to climb the stairs."

 

[NascentOxygen]

A 60.78kg student climbs 4.08m of stairs in 3.85s.

That would have to involve 4.08m measured vertically. I think that's the convention for stairs, hills, etc., and often reinforced by "climb", "rise", "ascend", or similar.