Work done on crate by the truck

AI Thread Summary
A truck accelerates a 65.4kg crate from rest to 61.3 km/h over 15.2 seconds, with non-constant acceleration. The initial calculation for work done was incorrectly based on a faulty speed conversion, leading to an erroneous result of 1.59e18. Upon correcting the conversion, the accurate work done on the crate is determined to be 9481 Joules. The equation W=Kf-Ki is confirmed as appropriate for calculating work in this context. The discussion emphasizes the importance of careful unit conversion and order of magnitude checks in physics calculations.
waldvocm
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A truck carrying a 65.4kg crate accelerates from rest to 61.3km/hr on a flat horizontal surface in 15.2s The acceleration was not constant. HOw much work was done on the crate by the truck?

W=Kf-Ki

W=(1/2*65.4kg*220680000^2)-0

W=1.59e18

Is this correct?
 
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Wow that's a very very fast truck. It's traveling at 0.7c!
Check your conversion of km/h to m/s again (I do have somewhat of an idea how you got that value). In any case, checking the order of magnitude of your answer often yields insights into possible careless errors in the working process.
 
opps! W=9481J Final answer!

Was I correct in choosing W=Kf-Ki for my equation in this problem?
 
waldvocm said:
Was I correct in choosing W=Kf-Ki for my equation in this problem?
Yes. Work done by an external force on a system increases the energy of the system.
 

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