How Many Kilocalories are Produced When a Truck Decelerates?

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Homework Help Overview

The problem involves a truck with a mass of 5850 kg that decelerates uniformly from an initial speed of 85.0 km/h to a final speed of 50.0 km/h. The task is to determine how many kilocalories are produced from the energy reduction, with the stipulation that 95.0% of this energy is converted into heat.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conversion of speeds from km/h to m/s and the calculation of kinetic energy using the formula Ek = 1/2mv². There is confusion regarding the correct algebraic manipulation of the kinetic energy equation, particularly in the order of operations and the treatment of initial and final speeds. Some participants question the subtraction of speeds and the correct approach to calculating the energy produced in kilocalories.

Discussion Status

The discussion includes attempts to clarify the algebra involved in calculating the change in kinetic energy. Some guidance has been provided regarding the correct order of operations and the need to convert energy from Joules to kilocalories. Participants are actively engaging with the problem, and there is a sense of progress as they refine their understanding of the calculations involved.

Contextual Notes

There is a noted mistake in the conversion of initial speed, which may affect the calculations. Participants are also navigating the constraints of homework rules that may limit the type of assistance they can provide.

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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.