How Many Kilocalories are Produced When a Truck Decelerates?

AI Thread Summary
A 5850kg truck decelerates from 85.0 km/h to 50.0 km/h, and the discussion focuses on calculating the kilocalories produced from this deceleration. The kinetic energy reduction is calculated using the formula Ek = 1/2mv², with the correct velocities converted to m/s. The initial confusion involved incorrect algebra, particularly in the order of operations for subtracting squared speeds. After clarifying the calculations and confirming that 95% of the energy reduction is converted into heat, the conversion from Joules to kilocalories was addressed. The discussion concludes with the participant successfully understanding the process to find the energy produced.
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Homework Statement


A 5850kg truck decelerates uniformly from 85.0km/h to 50.0 km/h. if 95.0% of the energy reduction is converted into heat, how many kilocalories are produced?



Homework Equations


kinetic energy (Ek)= 1/2mv2
1 Cal= 4.18J

The Attempt at a Solution


So first I converted 85.0km/h and 50.0km/h to m/s. 85.0km/h=26.4m/s, 50.0km/h=13.9m/s.
Since 26.4m/s is initial velocity and 13.9m/s is final velocity, so the uniform velocity is v1-vo,: 13.9m/s-23.6m/s=-9.7m/s. (Although I'm not quite sure it's right)

Then
Ek=1/2mv2
.5*(5850kg)(-9.7m/s)2=2.75*105
(.95%)(2.75*105)=2.61*105
so I converted it to cal which 1 cal= 4.18J. so answer came 6.25*104cal. and it was incorrect. someone help me?
 
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You just messed up the algebra. In particular this step:

.5*(5850kg)(-9.7m/s)2=2.75*105

You can't do that.

It should be:

0.5*(5850kg)*(26.4m/s)2 - 0.5*(5850kg)*(13.9m/s)2
 
hmm so speed not suppose to get subtracted?
sorry, my algebra is my most weakest spot.
 
Well the speeds are subtracted but you need to square them first before you subtract.
 
umm shouldn't final speed go first then initial speed go last on your solution?
how do you get the kcal part? I'm not sure what to do from there

Also I made mistake, initial speed should be 23.6m/s
 
Ok so firstly we are looking for a positive energy amount (they ask for energy produced) so we place the larger speed first.

The equation I gave you solves for the energy reduction of the truck in Joules. You need to figure out how much of that energy goes into heat and then convert from Joules to Kilocalories.
 
yay I got it. thank you.
 
No problem. Good job.
 

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