Calculating Work Done on Child & Sled Pulled by Force

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Homework Help Overview

The discussion revolves around calculating the total work done on a child and sled system being pulled by a force at an angle, while also considering the force of friction. The problem involves concepts from mechanics, specifically work and energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the work done by the pulling force and questions whether to include the work done against friction in the total work calculation. Participants discuss using the net work to find the final speed of the sled.

Discussion Status

Participants are actively engaging with the problem, confirming the need to calculate work done by both the pulling force and friction. Some guidance has been offered regarding the use of net work to find the final speed, but no consensus has been reached on the complete solution.

Contextual Notes

The original poster expresses a lack of prior physics experience and seeks reassurance in their understanding of the concepts involved.

fa08ti
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A child on a sled (having a combined mass of 47.0 kg) is pulled by a force directed along a rope that makes a 45 degree angle with the horizontal axis. the force exerted on the rope is 100.00 N. the force of friction acting on the sled is 30.0 N. if the child is pulled a distance of 10.0 m along a level field, determine the total work done on the child and the sled.

Attempt:
Given:
d: 10.0 m, F:100.0 N (45 degrees), Ff: 30.0 N (180 Degrees)

W=F(costheta) . deltad
W= 100.00 N X cos45 X 10.0m
W=707J
that would be just for the rope

would i have to do the same calculation using the Ff now? then add both numbers to get total work done?

It also asks for the final speed at the end of 10.0 m

HELP PLEASE!
 
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fa08ti said:
would i have to do the same calculation using the Ff now? then add both numbers to get total work done?
Yes.

It also asks for the final speed at the end of 10.0 m
Make use of the net work done that you've calculated.
 
ok so for the speed, i'd use
W= (1/2)mv^2 and just rearrange it
 
Last edited:
fa08ti said:
ok so for the speed, i'd use
W= (1/2)mv^2 and just rearrange it
Good!
 
thanks sooooo much. I've never taken physics before and I'm finding that while i understand most concepts, I tend to need a lot of reassurance
 

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