What is the energy lost to friction?

AI Thread Summary
The discussion revolves around calculating the energy lost to friction for a block being dragged across a surface. The user initially believes that the energy lost is equal to the negative work done by the frictional force, estimating it to be around -215 J. However, confusion arises regarding the angle used in the calculations, as the correct angle should reflect the force's direction rather than 180 degrees. Participants emphasize the importance of accurately determining the normal force and the gravitational component acting on the block. Clarification on these points is necessary to resolve the user's misunderstanding and correctly compute the energy lost to friction.
Kumar9
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The question is: A 15.7 kg block is dragged over a horizontal surface by a 68.7N force acting 17 degrees to the horizontal. The block is displaced 4.65m, and the kinetic friction coefficient is 0.3. What is the energy lost to friction?

My thinking is that it would be equal to the negative work done by the frictional force (E taken out of the system), but the computer does not recognize this as correct. What am I missing?
 
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I need to have this figured out soon; if someone could help I would greatly appreciate it.
 
Kumar9 said:
The question is: A 15.7 kg block is dragged over a horizontal surface by a 68.7N force acting 17 degrees to the horizontal. The block is displaced 4.65m, and the kinetic friction coefficient is 0.3. What is the energy lost to friction?
My thinking is that it would be equal to the negative work done by the frictional force (E taken out of the system), but the computer does not recognize this as correct. What am I missing?
Show us your work. We can't help you otherwise. What is the force of friction?

AM
 
This is what I have so far:
W=(mgUk)(d)cos180
Where Uk is the coefficient. Plugging in the appropriate values, I get a value of ~-215 J, the E lost due to friction. It seems right, but clearly I'm missing something.
 
Kumar9 said:
This is what I have so far:
W=(mgUk)(d)cos180
Where Uk is the coefficient. Plugging in the appropriate values, I get a value of ~-215 J, the E lost due to friction. It seems right, but clearly I'm missing something.
I don't understand why you are using 180 degrees. The angle is 17 degrees from horizontal. The normal force perpendicular to the surface is equal and opposite to the component of gravity in that direction. What is the magnitude of that component of gravity? (Hint it is a little less than the weight).

AM
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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