Work done by gravity on a car rolling down a hill

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The discussion centers on calculating work done by gravity on a car rolling down a hill, with a focus on the formula E = F x cos(θ). A participant mentions obtaining 243 kJ but questions the values used for force (F), distance (x), and angle (θ). The conversation suggests that using the work-energy principle, W = ΔE_P, is more appropriate for this scenario than the traditional formula. Additionally, there is a query about the specific distance represented by 50 cos(8°) m in the context of the problem. The thread emphasizes the importance of correctly applying the relevant formulas to accurately determine the work done by gravity.
physicsmaster123
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I tried E =Fxcos0 but only ended up with 243kJ
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What values of ##F##, ##x##, and ##\theta## did you use and why?
 
Hint: What is the textbook definition of ##\theta## in the formula you stated?
 
Last edited:
physicsmaster123 said:
I tried E =Fxcos0 but only ended up with 243kJ
What distance in the diagram is ##50\cos(8°)##m?
 
Specific to this exercise, ##W=\Delta E_P## is a better fit than ##W=Fd\cos \theta##
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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