Work/Energy: Find Initial Speed of 0.50kg Ball Down Incline

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Homework Help Overview

The problem involves a 0.50 kg ball projected down an incline from a height of 7.5 m, with friction producing 10.7 J of heat energy. The ball subsequently reaches a height of 13.0 m after leaving the incline. The objective is to determine the initial speed of the ball at the starting position, while ignoring air resistance.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, particularly in relation to the work done by friction. There is uncertainty regarding whether to treat the energy lost due to friction as a positive or negative value in calculations. Some participants suggest rephrasing the energy equation to clarify the relationship between initial and final energies.

Discussion Status

The discussion is ongoing, with participants exploring the implications of energy loss due to friction and how it affects the energy balance in the system. Some guidance has been offered regarding the formulation of the energy equation, but no consensus has been reached on the correct sign to apply to the heat energy produced by friction.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to show work in a specific format as preferred by the instructor. There is also a focus on understanding the implications of energy conservation in the presence of non-conservative forces like friction.

Inertialforce
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Homework Statement


A 0.50 kg ball starting from position A which is 7.5m above the ground, is projected down an incline as shown. Friction produces 10.7J of heat energy.

The ball leaves the incline at position B traveling straight upward and reaches a height of 13.0m above the floor before falling back down.

What was the initial speed, Vi, at position A (ignore air resistance)?

This picture link that I have posted with this question is the exact same picture that I have on my question sheet except the height value at B is 13.0m not 4.00m and the height value at A is 7.5m not 3.00m.
http://qaboard.cramster.com/answer-board/image/5b5efcf498963c65f5517ad6687243f2.jpg


Homework Equations


ΔE = heat

(I know that we can use the equation Wnc = ΔE, but our teacher would prefer it if we used this method to show how we got the answer)

The Attempt at a Solution


My question is that, I know that it says friction/heat is produced but what sign do I apply to it (positive or negative)? I put it into the equation as a positive value because it says friction is "produced", but anyway could someone tell me how you would know if it is positive or negative and if my calculations are okay?

Since it says that friction produces 10.7J of heat energy, I know that work is not conserved in this system. Therefore I used the equation:

ΔE = heat
ΔEp + ΔEk = heat
(mghf - mghi) + ((1/2)mvf(squared) - (1/2)mvi(squared)) = heat
mghf - mghi - (1/2)mvi(squared) = heat
(0.50)(9.80)(13.0) - (0.50(9.80)(7.5) -(1/2)(0.50)vi(squared) = 10.7
63.7 - 36.75 -(1/2)(0.50)vi(squared) = 10.7
26.95 -10.7 = (1/2)(0.50)vi(squared)
(2)(16.25) = (0.50)vi(squared)
32.5/0.50 = vi(squared)
√65 = √(vi(squared)
8.062257748 = vi
8.06 m/s = vi
 
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Inertialforce said:
My question is that, I know that it says friction/heat is produced but what sign do I apply to it (positive or negative)? I put it into the equation as a positive value because it says friction is "produced", but anyway could someone tell me how you would know if it is positive or negative and if my calculations are okay?

Hi Inertialforce! :smile:

Write the equation in the form:

energy before = energy after ± …

you know it's lost energy, so which side must be larger? :wink:

(sorry, i didn't chcek your equations, they're too difficult to read :redface:)
 
tiny-tim said:
Hi Inertialforce! :smile:

Write the equation in the form:

energy before = energy after ± …

you know it's lost energy, so which side must be larger? :wink:

(sorry, i didn't chcek your equations, they're too difficult to read :redface:)


so, energy is lost therefore the right side has to equal a negative number? Which makes the left side larger meaning that either the change in potential energy is larger than the change in kinetic energy or vice versa. Would this be correct?
 
Inertialforce said:
so, energy is lost therefore the right side has to equal a negative number? Which makes the left side larger meaning that either the change in potential energy is larger than the change in kinetic energy or vice versa. Would this be correct?

I think your confusing yourself by looking at it that way …

you're certainly confusing me :redface:

Just say KE + PE before = KE + PE after + energy lost. :smile:
 

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