Finding the Change in Kinetic Energy of a Box on a Rough, Horizontal Floor

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Homework Help Overview

The problem involves a box being pushed along a rough, horizontal floor, with a focus on determining the change in kinetic energy. The scenario includes specific values for mass, applied force, distance, and friction coefficient.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the work-energy theorem and its application to the problem. Questions arise regarding how to find the change in kinetic energy without knowing the final velocity. There is also an exploration of net force and work done against friction.

Discussion Status

Participants are actively engaging with the concepts, confirming each other's understanding of the work-energy theorem. There is a collaborative effort to clarify the relationship between work done and changes in kinetic energy, with some guidance provided on calculating net force.

Contextual Notes

There is an emphasis on the need for velocity to determine kinetic energy, which remains a point of inquiry. The discussion reflects a collaborative homework environment, with participants working together to understand the problem better.

scurry18
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Homework Statement


A 40.0kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. The coefficient of friction between box and floor is .300.
Find the change in kinetic energy of the box.


Homework Equations


(delta)K=K(final)-K(initial)
K(final)=1/2mv^2


The Attempt at a Solution


Work=650. J
(change)Internal Energy=588 J
Fk=118

How do you find the change in Kinetic energy without the velocity?
 
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What does the work-energy theorem state?
 
When work is done on a system and the only change in the system is in its speed,
the work done by the net force equals the change in kinetic energy of the system.

Wnet = Kf - Ki = deltaK

In this situation Ki = 0, Kf = 1/2mv^2

Right?
This is a classmate of mine and we are working on this together.
Kevin
 
That's correct. So you have the distance over which the force acts. What is the net force in this situation?
 
It is the Work force of 650J - the frictional force of 588J to get 62J?
Kevin
 
That looks right to me.
 
Now we will try and work on her other one with the skier that she posted and we
should be done.
Thanks for the help,
Kevin
 

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