How much energy is dissipated as heat via friction?

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SUMMARY

The discussion focuses on calculating the work done and energy dissipated as heat due to friction when dragging a 55 kg box with a constant force of 100N at a 15-degree angle over a rough surface. The net work done is calculated using the formula W = Fd, resulting in 129.6 Joules. The energy dissipated as heat due to friction is determined by the friction force of 75N over a distance of 6m, yielding 450 Joules. The final velocity of the box can be calculated using the work-energy theorem, considering the kinetic energy formula E = 1/2mv².

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dangish
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2. Hubert drags a 55 kg box over a rough surface of a dock with a constant force of 100N acting at an angle of 15 degrees to the horizontal. The box moves in a straight line for a distance of 6m. Friction acts at an opposite direction to the block and it is 75N. Calculate the work done to move the box? How much energy is dissipated as heat via friction? What is the final velocity of the box?

Ok so i started W=Fd

Would the force be the forces in the x direction? ie: 100cos15 - 75N ?

Which would give W = (100cos15 - 75) x 6 = 129.6Joules ??

Would that also be the energy dissipated as heat?

If so I could then use that to find the final velocity with E=1/2mv^2
 
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dangish said:
2. Hubert drags a 55 kg box over a rough surface of a dock with a constant force of 100N acting at an angle of 15 degrees to the horizontal. The box moves in a straight line for a distance of 6m. Friction acts at an opposite direction to the block and it is 75N. Calculate the work done to move the box? How much energy is dissipated as heat via friction? What is the final velocity of the box?

Ok so i started W=Fd

Would the force be the forces in the x direction? ie: 100cos15 - 75N ?
There are forces in the x and y direction, but you are only intersted in the x direction for this problem, since the forces in the y direction do not do any work. you have correctly calculated the net force in the x direction.
Which would give W = (100cos15 - 75) x 6 = 129.6Joules ??
yes, that is the net (or total) work done by all forces.
Would that also be the energy dissipated as heat?
the problem asks for the energy dissipated as heat by the friction force.
If so I could then use that to find the final velocity with E=1/2mv^2
Use what to find the final velocity (please clarify)?
 


So to find the energy dissipated by friction I should go W= 75 x 6 ??
 


dangish said:
So to find the energy dissipated by friction I should go W= 75 x 6 ??
Yes!
 


Then what about the final velocity? I know I have to use E=1/2mv^2

But It doesn't seem right to use the energy dissipated by the friction force
 


dangish said:
Then what about the final velocity? I know I have to use E=1/2mv^2

But It doesn't seem right to use the energy dissipated by the friction force
You are correct in that it doesn't seem right to just use the dissipated energy from friction, so instead, use the work-energy theorem to solve for the final KE.
 

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