How can I solve this Work Energy Theorem problem for finding height?

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Homework Help Overview

The discussion revolves around a Work Energy Theorem problem where participants are attempting to find the height of a truck's ascent. The original poster expresses confusion about using the equation mgh = ΔK.E. and mentions a specific height provided by a teacher.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between work done by forces and changes in mechanical energy, questioning the roles of conservative and non-conservative forces. Some suggest considering the truck and Earth as a system to account for external forces.

Discussion Status

There is ongoing exploration of the problem with various approaches being discussed. Some participants have offered guidance on how to set up equations involving work and energy, while others are still seeking clarity on specific aspects of the problem.

Contextual Notes

Participants note the importance of accounting for friction as a non-conservative force and discuss the implications of constant speed on kinetic energy changes. There is also mention of needing to clarify the definitions and roles of work done by different forces in the context of the problem.

TheRedDevil18
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Homework Statement


I do not know how to find the height. I am thinking of using mgh=delta k but I am not getting the correct answer. I know the answer is 6.28m because my teacher gave it to me.
Below is the picture and question:
2hcm8uh.jpg


Homework Equations


mgh = delta k

The Attempt at a Solution


ep=ek
mgh=785000
98000h=785000
h = 8.01m
 
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TheRedDevil18 said:
I do not know how to find the height. I am thinking of using mgh=delta k but I am not getting the correct answer.
Since the truck moves at constant speed, ΔKE = 0. So that won't work.

Instead, consider the work done by the engine and where that work goes.
 
Take truck as the system. What force acts during its ascension ? How many are non conservative ? What is work done by non conservative force ?

Negative work done by non conservative force= Change in truck's total mechanical energy.
 
sankalpmittal said:
Take truck as the system. What force acts during its ascension ? How many are non conservative ? What is work done by non conservative force ?

Negative work done by non conservative force= Change in truck's total mechanical energy.

The non conservative force will be the friction right? so if I set up the equation it will be like this
-85000=k1+u1
 
TheRedDevil18 said:
The non conservative force will be the friction right? so if I set up the equation it will be like this
-85000=k1+u1

I should have told you to take truck+earth as the system to account for external forces.

Yes. Now total change in mechanical energy is change in kinetic energy plus change in potential energy. As change in kinetic energy is zero, so what is the change in potential energy? It will be better to account for minus sign to gravity.
 
mgh, Cant work it out, need the height.
 
TheRedDevil18 said:
mgh, Cant work it out, need the height.

Wfriction = M.Ef-M.Ei

M.E. is mechanical energy..

Wfriction = ΔK.E + ΔP.E.

As ΔK.E=0 , so

ΔPE=mgh

Now you know, Wfriction= -85000 J.. Can you work it out from here ?
 
TheRedDevil18 said:
mgh, Cant work it out, need the height.
The height is what you are supposed to find! Just call it h and you will solve for it.
 
Okay I think I got the answer:
Fapp-Ffric=mgh
7*10^5-8.5*10^4=10000*9.8*h
h = 6.28m

Correct?
 
  • #10
TheRedDevil18 said:
Okay I think I got the answer:
Fapp-Ffric=mgh
7*10^5-8.5*10^4=10000*9.8*h
h = 6.28m

Correct?
Yes, correct. (Where Fapp is the work done by the applied force, the engine in this case. Better to call it Wapp, right? And the work done against friction would be Wfric.)
 
  • #11
Yeah, sorry its work not force. Anyway,for 3.2 what do they mean by the instantaneous power delivered by the truck. I know P=w/t and I got the time to be 4.14s using t=d/s but what should I use for work, must I subtract Wapp-Wfrict and put it into the equation?
 
  • #12
TheRedDevil18 said:
Anyway,for 3.2 what do they mean by the instantaneous power delivered by the truck. I know P=w/t and I got the time to be 4.14s using t=d/s
Good.
but what should I use for work, must I subtract Wapp-Wfrict and put it into the equation?
To find the power delivered by the engine, just use the work done by the engine.
 
  • #13
Okay so,
p=w/t
= 7*10^5/4.14
= 169082.13 W

Correct?
 
  • #14
TheRedDevil18 said:
Okay so,
p=w/t
= 7*10^5/4.14
= 169082.13 W

Correct?
Good. (I suggest rounding off your final answer and using exponential notation.)
 
  • #15
Will do, and thanks for your help
 

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