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

In summary, the homework statement is asking for someone to solve for the height of a truck that has been raised by an engine. The student is having trouble getting the answer, so they ask for help. The help they receive is to use an equation that accounts for external forces, friction, and work done by the engine. The student realizes that they need to find the height of the truck and so they use an equation to calculate it.
  • #1
TheRedDevil18
408
1

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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  • #2
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.
 
  • #3
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.
 
  • #4
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
 
  • #5
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.
 
  • #6
mgh, Cant work it out, need the height.
 
  • #7
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 ?
 
  • #8
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.
 
  • #9
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
 

1. What is the Work Energy Theorem?

The Work Energy Theorem is a fundamental principle in physics that states that the work done on an object is equal to the change in its kinetic energy. This means that when a force is applied to an object, it will either speed up or slow down depending on the direction of the force.

2. How is the Work Energy Theorem expressed mathematically?

The Work Energy Theorem is expressed as W = ΔKE, where W represents the work done on an object and ΔKE represents the change in kinetic energy. This equation can also be written as W = Fd, where F is the force applied to the object and d is the distance over which the force is applied.

3. What are the units of work and kinetic energy?

The units of work are joules (J), which is equivalent to a Newton-meter (N·m). The units of kinetic energy are also joules (J). This makes sense since work and kinetic energy are directly related through the Work Energy Theorem.

4. How is the Work Energy Theorem applied to real-world problems?

The Work Energy Theorem can be applied to various real-world problems, such as calculating the work done by a car's engine to accelerate it to a certain speed, or the work done by a person lifting a box off the ground. By using the equation W = ΔKE, the work can be calculated by determining the change in kinetic energy of the object.

5. Are there any limitations to the Work Energy Theorem?

While the Work Energy Theorem is a useful principle for solving many physics problems, it does have limitations. It assumes that there are no other forces acting on the object besides the applied force, and that the force is constant over the entire distance. In reality, there may be other forces at play and the applied force may vary. In these cases, the Work Energy Theorem may not provide an accurate solution.

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