What is the force required to push a block of ice up an incline?

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Homework Help Overview

The discussion revolves around determining the force required to push a 500N block of ice up a 3m incline while neglecting friction. Participants are exploring the relationship between the incline's dimensions and the force needed to overcome gravity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to clarify the dimensions of the incline and the relationship between the distance along the incline and the vertical height. Questions about the angle of the incline and its implications for the force required are raised.

Discussion Status

The discussion is active, with participants exploring various interpretations of the incline's dimensions and the necessary calculations. Some guidance has been offered regarding the need for a diagram and the implications of the incline's angle on the force required.

Contextual Notes

There is some ambiguity regarding the definitions of the incline's height and distance, which affects the calculations. Participants are also considering the implications of gravitational force in their reasoning.

polishguy1234
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how much force is needed...

Homework Statement



How much force is needed to push a 500N block of ice 6m up a 3m incline. and neglect friction

Homework Equations


work =fd

The Attempt at a Solution

 
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polishguy1234 said:

Homework Statement



How much force is needed to push a 500N block of ice 6m up a 3m incline. and neglect friction

Homework Equations


work =fd

The Attempt at a Solution


Welcome to PF.

Any ideas on how you can push something 6m up a 3m incline?
 


thanks. and no i don't know how to do this. I am trying to learn though.
 


So is 3m the height of the incline? And 6m the distance of the incline?
 


Yeah
 


So what is that angle?

Knowing the angle then you can determine the force against gravity needed to push the block can't you?
 


If you use the formula W=fd:

500N(weight of an object can also be considered the force) X 6m ( the distance) = 3000 J


I hope this helped
 


Be careful Jenn and make a diagram of this situation. The object is moving 6m but does it take 500N or a force equal to the weight to move it the 6 m?

The problem appears to indicate that the path of the mass is 6m and the vertical height is 3m. It's not clear. That, along with the weight of the object is adequate to calculate the force needed to move it up the ramp.
 


Well 3 m height and 6m distance would be like a road inclination of 50%(per cent, not really percentage but per cent as in hundred) or 50 m height gain every 100m distance.
So 50% inclination would resemble like 26.57 degree according to the Grade (slope) page on wikipedia.
So now you know 6 m distance, 3 m height with a 26.57 degree inclination.
 
  • #10


Conservative force of gravity.
For conservative force, the work is the same along all paths joining two given points.
 

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