How to find work from mass, friction given?

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Homework Help Overview

The problem involves calculating the work done by a child dragging a box across two different surfaces with varying coefficients of friction. The scenario includes a box being pulled horizontally over a distance of 14.0 m on grass and 36 m on a sidewalk, with given mass and friction values.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work in relation to force and distance, questioning how to determine the frictional force and whether to combine distances from different surfaces. There are also inquiries about the relevance of additional information provided in the problem.

Discussion Status

Participants have explored various aspects of the problem, including the calculation of forces and the application of the work formula. Some have provided hints and guidance on how to approach the calculations, while others have raised questions about the assumptions and setup of the problem.

Contextual Notes

There is a discussion about the different coefficients of friction for each surface and how they affect the calculations. The relevance of additional parameters, such as time, is also questioned but remains unclear in its impact on the solution.

Sneakatone
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a Child drags a 25-kg box at a constant speed across a lawn for 14.0 m and along a sidewalk for 36 m; the coefficient of friction is 0.25 for the first part of the trip and 0.55 for the second. If the child always pulls horizontally, how much work does the child do on the box?I know that Work= force*distance but how would I find these variables?
idk if this works but I can do 25kg * 9.81=245 as force
 
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Hint: The child is doing work by pulling against the frictional force on the box. Do you know how to calculate this force?
 
That is a relevant force yes, but it is acting only in y, and is canceled out by the normal force.

First. Draw a free body diagram. What is the free body? What forces are acting on this free body?

Hint: How much force does it take to overcome the friction in each individual part of the trip?
 
245 n *0.25=61.25n
 
Good, now what?
 
we have to find the distance , I don't suppose we add 14 + 36=50m for the distance, do we?
 
is the 55 sec relavent in this problem?
 
Sneakatone said:
we have to find the distance , I don't suppose we add 14 + 36=50m for the distance, do we?
Not quite, you have two different surfaces. Each with a different coeffecient of friction.
Sneakatone said:
is the 55 sec relavent in this problem?
I believe that is .55 for the coefficient of friction for the sidewalk.
 
1st part)25*.25*14=87.5J
2nd part) 25*.55*36=495J
 
  • #10
Sneakatone said:
1st part)25*.25*14=87.5J
2nd part) 25*.55*36=495J
You're missing something. What are you multiplying? You found the force needed to move across the grass here:
Sneakatone said:
245 n *0.25=61.25n
Yet, you just threw that value to the wind and multiplied 3 values together.

You know that W=FDcosθ. And you know the the displacement. So what now?

*where theta is the angle between the Applied force and the displacement. (In this case 0, cos0=1.)*
 
  • #11
245N*0.25*14m=857.5J
245N*0.55*36m=4851
 
  • #12
I added them both and it was correct, Thank you!
 
  • #13
Nice! Just keep in mind that Force and Displacement are vectors. Work isn't just F times D. It's F times D times the cos of the angle between them. In nice situations the angle is 0. But in some cases, the there will be an angle between them, and then it can get a little tricky!
 

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