Relatively simple mechanics problem

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A physics problem involves calculating the height of an incline where an object is pushed up and down at different speeds. The person pushes the object up at 15 km/hr and down at 30 km/hr over a 4 km distance. Discussions include the application of forces, acceleration, and the role of kinetic and potential energy in the problem. Participants explore various equations and concepts, including the conservation of energy and the effects of constant force. The conversation emphasizes the need for clarity in assumptions and calculations to solve for the incline's height effectively.
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Homework Statement



A person pushes an object along an incline at an average speed of 15km/hr.After reaching the top, he pushes the object back from top to bottom, where the average speed is 30km/hr.if the length along which he pushes( hypotenuse of the inclined plane) is 4km and the force is constant, Find the height of the inclined plane.THE QUESTION CAN BE EASILY GRASPED BY LOOKING AT THE REFERENCE PIC.

Homework Equations



Don't know which one to use.

The Attempt at a Solution


Ok, so frankly i got nothing.but, since the force is constant and the speeds are different, i guess we can put F1=mv1/t1=mv12/s?
And F2=F1+mgsinΘ? That's all i got.please help
 

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Why do you think he has to push to get it back down?
 
I know, he won't HAVE to push(the object would just slide,F=mgsinθ), but that's what the question says.BESIDES, since he applied the same force, v should have been same(in case of straight plane), but v2 is faster because(PROBABLY) F2=F1+mgsinθ
 
Paradox101 said:
I know, he won't HAVE to push(the object would just slide,F=mgsinθ), but that's what the question says.BESIDES, since he applied the same force, v should have been same(in case of straight plane), but v2 is faster because(PROBABLY) F2=F1+mgsinθ
It doesn't say whether there is friction, so I considered there might be.
Ignoring that for now, if he applies force F going up, what is the acceleration?
If the distance to the top is s, what is the average speed?
 
haruspex said:
It doesn't say whether there is friction, so I considered there might be.
Ignoring that for now, if he applies force F going up, what is the acceleration?
If the distance to the top is s, what is the average speed?
Neglect friction.sorry for not adding that.And,just for convenience, I converted the distance and time units(km and hr) to m and s.Now,average speed to the top is given.
So, time=4000m/[(15*5/18)m/s]=960 seconds.
⇒a1=v/t=(15*5/18)/960=0.00434m/s2
What next? do i find the mass(if so,how and why?)?
 
Paradox101 said:
Neglect friction.sorry for not adding that.And,just for convenience, I converted the distance and time units(km and hr) to m and s.Now,average speed to the top is given.
So, time=4000m/[(15*5/18)m/s]=960 seconds.
⇒a1=v/t=(15*5/18)/960=0.00434m/s2
What next? do i find the mass(if so,how and why?)?
Please explain your calculation of the acceleration. What did you divide by what, exactly?
Then write out the appropriate equation for the forces giving rise to this acceleration.
It's late here, back tomorrow.
 
haruspex said:
It's late here, back tomorrow.
LOL, sorry, didn't realize that(Australia-----diff. time zone*duh*)
Is it wrong though? I mean, s=v/t, where v=15km/hr=15*5/18 m/s, and s = 4000m, so t= s/v=960s(used the calc.).
and, a=v/t=(15*5/18)/960=0.00434m/s^2.
 
Paradox101 said:
LOL, sorry, didn't realize that(Australia-----diff. time zone*duh*)
Is it wrong though? I mean, s=v/t, where v=15km/hr=15*5/18 m/s, and s = 4000m, so t= s/v=960s(used the calc.).
and, a=v/t=(15*5/18)/960=0.00434m/s^2.
I meant that last step. Where you have applied v/t, how would you define the velocity you used? What is the definition of v in a=v/t? What velocity does v represent?
 
haruspex said:
I meant that last step. Where you have applied v/t, how would you define the velocity you used? What is the definition of v in a=v/t? What velocity does v represent?
so,i have been trying something, is it possible that i could first find t1=960s,t2=480s, then the average speeds are 0.24m/s and 0.12m/s, and 4000F=mgh?
 
  • #10
Paradox101 said:
4000F=mgh?
what other work has F done?
 
  • #11
don't know...moved the object against the component of the gravitational force?
 
  • #12
CORRECTION: the average speeds are------- 4000/960=4.1666...m/s and 4000/480=8.3333...m/s.
 
  • #13
Paradox101 said:
don't know...moved the object against the component of the gravitational force?
As you calculated, the object accelerates up the slope. Since the force is constant, that acceleration is constant.
The question as stated could be clearer. You need to assume that the object starts from rest in each direction, and is somehow brought to a sudden stop at the end in each case. So, what other form of energy do you need to consider?
 
  • #14
haruspex said:
As you calculated, the object accelerates up the slope. Since the force is constant, that acceleration is constant.
The question as stated could be clearer. You need to assume that the object starts from rest in each direction, and is somehow brought to a sudden stop at the end in each case. So, what other form of energy do you need to consider?
Kinetic?
 
  • #15
Paradox101 said:
Kinetic?
Yes.
 
  • #16
haruspex said:
Yes.
Are you implying that 4000F= mv2/2? If so, which speed is v?
 
  • #17
Paradox101 said:
Are you implying that 4000F= mv2/2? If so, which speed is v?
No, I'm saying that in both directions the object starts with no KE but finishes with KE.
Going uphill, it also gains PE, but loses it downhill.
You can work out the final speed in each case from the average speed and the fact that forces are constant.
 
  • #18
Ok final speed in first case will be twice of average speed( because initial speed is zero)=25/3m/s.so,KE1=mv2/2=625m/18.
And final speed in second case will be 2×30×5/18=50/3m/s.
Now what?the 2 KEs are different.I'm not sure, but it won't be W=KE2-KE1, right?because that's a different thing.
 
  • #19
Paradox101 said:
Ok final speed in first case will be twice of average speed( because initial speed is zero)=25/3m/s.so,KE1=mv2/2=625m/18.
And final speed in second case will be 2×30×5/18=50/3m/s.
Now what?
So correct your energy equation in post #9.
Paradox101 said:
the 2 KEs are different.I'm not sure, but it won't be W=KE2-KE1, right?because that's a different thing.
The two phases, uphill and downhill, are completely separate. There is no carry over of KE from the first to the second. Why would you take the difference of the two KEs?
 
  • #20
But what should I do next?How do I relate these KEs with the height of the incline?
 
  • #21
Paradox101 said:
But what should I do next?How do I relate these KEs with the height of the incline?
Please post a corrected energy conservation equation for, say, the uphill stage. You know the total work done, and you can calculate how much of that went into KE. Where did the rest of the work go?
 
  • #22
An alternative solution : Consider two unknowns , force by man , say F , and angle θ .

Use kinematics formula - Si = 0.5*Aata2 ,
and Si = 0.5*Adtd2 ,

where 'A ' refers to acceleration , ' a ' refers to ascent , ' d ' to descent , and Si to displacement along incline .
You know pretty much everything . Make two equations , and solve .

All assumptions made in previous posts are considered valid .

Hope this helps .
 
  • #23
Qwertywerty said:
An alternative solution : Consider two unknowns , force by man , say F , and angle θ .

Use kinematics formula - Si = 0.5*Aata2 ,
and Si = 0.5*Adtd2 ,

where 'A ' refers to acceleration , ' a ' refers to ascent , ' d ' to descent , and Si to displacement along incline .
You know pretty much everything . Make two equations , and solve .

All assumptions made in previous posts are considered valid .

Hope this helps .
The thread started out with forces and accelerations, but at post #9 Paradox seemed to want to switch to using energy. I don't care as long as we make progress.
 
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