Calculating Power for Pushing a Lawnmower at Constant Speed

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Homework Help Overview

The problem involves calculating the power required to push a 12 kg lawnmower at a constant speed of 1.2 m/s, considering the angle of the handle and the coefficient of rolling friction. The original poster attempts to apply the power equation but expresses uncertainty in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the lawnmower, including the normal force and friction. There are questions about the correct application of the power equation and the components of forces in both the x and y directions.

Discussion Status

Some participants have provided hints regarding the need for a free-body diagram and the interpretation of forces, while others are seeking clarification on the calculations and assumptions made by the original poster. The discussion is ongoing with various interpretations being explored.

Contextual Notes

The original poster has indicated a misunderstanding regarding the units of power and the relationship between force and motion, which may affect their calculations. There is also a mention of needing to show more steps in their reasoning.

ceday
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Homework Statement



A gardener pushes a 12 kg lawnmower whose handle is tilted up 37 deg above horizontal. The lawnmower's coefficient of rolling friction is 0.15 How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2 m/s Assume his push is parallel to the handle.

Homework Equations



power = force x velocity

The Attempt at a Solution



I know the answer is 24 joules but i can't get there

the equation I used to find the force was F= [tex]\mu[/tex]kmg/ [tex]\mu[/tex]ksin[tex]\theta[/tex] + cos[tex]\theta[/tex] +sin[tex]\theta[/tex]

i know its not right and i can't find the error

thanks!
 
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Where is your power equation? Show us how you're calculating power and we might be able to help you.

BTW, Joules is a measure of energy, not power... (are you really looking for power?) - In any case, I think you need to show more information (more steps...)

-Kerry
 
Okay

For the Y direction forces, I was using y= -mg+n where n= mg + fsin[tex]\theta[/tex]?

for x force ,
x=-fk + fcos[tex]\theta[/tex] then,
x=-[tex]\mu[/tex]kn + fcos[tex]\theta[/tex] where again n= mg + fsin[tex]\theta[/tex]?



I think the term that I am getting hung up on is the normal force
the lawn mower isn't noving up or down so shouldn't the y force = 0?
 
Try drawing a free-body diagram of the lawn mower. The forces acting on the lawn mower are allowed to have components in the y-direction. AND what you said is true... the lawn mower is not moving vertically. Is this enough of a hint?

If you still need help with this step, post your free-body diagram (don't forget to include accelerations).

-Kerry
 

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