(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

kinetic energy (Ek)= 1/2mv^{2}

1 Cal= 4.18J

3. 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=1/2mv^{2}

.5*(5850kg)(-9.7m/s)^{2}=2.75*10^{5}

(.95%)(2.75*10^{5})=2.61*10^{5}

so I converted it to cal which 1 cal= 4.18J. so answer came 6.25*10^{4}cal. and it was incorrect. someone help me?

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# Homework Help: Mechanical equivalent of heat

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