# Work & KE

1. Sep 7, 2005

### rueberry

How much work is required to stop a 1000 kg car traveling at 100 km/hr?

I was going to use KE=1/2mv^2 , since the work should equal the KE, assuming no friction.

I have: =1/2(1000 kg)(100km/hr)^2

I was using dimensional analysis to make the conversion to m/s, so I end up with J, but I think I'm doing something wrong in the conversion.

(1000 km/hr)(1000 m/1 km)(3600 sec/1 hr)

2. Sep 7, 2005

### Andrew Mason

To get Joules, the conversion factor from km/hr to m/s is: 1000/3600 = .277. 1 km/hr = .277 m/sec; 100 km/hr = 27.7 m/sec.

AM

3. Sep 7, 2005

### rueberry

When I work that problem out, I get 383645J, but the answer in the book says 386,000J. Am I doing something wrong? If I use 27.7 m/sec in the equation, I get

1/2(1000kg)(27.7m/sec)=383645

4. Sep 7, 2005

### whkoh

The book is using the ratio 100*1000/3600 [=27 7/9] and not the approximation 27.7 m/s.