Power up and incline with friction

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Homework Help Overview

The problem involves a 1400 kg block of granite being pulled up an incline at a constant speed, with given parameters such as the angle of inclination and the coefficient of kinetic friction. The objective is to determine the power required by the winch to maintain this motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equation used for tension and question its correctness. There are attempts to clarify whether the issue lies in the equation itself or in the values substituted. One participant expresses uncertainty about the equations needed and seeks hints.

Discussion Status

The discussion is ongoing, with participants exploring the validity of the equations and values used. Some guidance has been offered regarding the relationship between tension, frictional resistance, and weight down the incline, indicating a potential direction for further exploration.

Contextual Notes

One participant mentions a lack of access to their notebook, which may limit their ability to reference equations during the discussion.

Kajayacht
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Homework Statement


A 1400 kg block of granite is pulled up an incline that has an angle of inclination θ = 29 ° with a constant speed of 1.13 m/s by a steam winch (see Figure). The coefficient of kinetic friction between the block and the incline is 0.12. How much power must be supplied by the winch?


Homework Equations


T= uW/cos(theta)+u*sin(theta)
P=Fv


The Attempt at a Solution


T= .12(1400*9.8)/cos(29)+.12*sin(29)
T= 1765.0 N

P= FV
P= 1765*1.13
P= 1994.5 W
 
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Kajayacht said:

Homework Statement


A 1400 kg block of granite is pulled up an incline that has an angle of inclination θ = 29 ° with a constant speed of 1.13 m/s by a steam winch (see Figure). The coefficient of kinetic friction between the block and the incline is 0.12. How much power must be supplied by the winch?


Homework Equations


T= uW/cos(theta)+u*sin(theta)
P=Fv

3. The Attempt at a Solution
T= .12(1400*9.8)/cos(29)+.12*sin(29)
T= 1765.0 N

P= FV
P= 1765*1.13
P= 1994.5 W


That equation does not look correct
 
wait do you mean the equation itself or I put in the values wrong?
 
Kajayacht said:
wait do you mean the equation itself or I put in the values wrong?

That equation is incorrect.
 
Figured that much, but I was hopeful.

Could you give me a hint or something, I think I might now the equations I need but my notebook is at my dorm right now and my next class is about to start.
 
Kajayacht said:
Figured that much, but I was hopeful.

Could you give me a hint or something, I think I might now the equations I need but my notebook is at my dorm right now and my next class is about to start.

Tension is going to equal the frictional resistance and the weight down the incline.

You don't need your notebook.
 

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