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Work & KE

  • Thread starter rueberry
  • Start date
5
0
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)
 

Answers and Replies

Andrew Mason
Science Advisor
Homework Helper
7,549
318
rueberry said:
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)
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
 
5
0
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
 
29
0
The book is using the ratio 100*1000/3600 [=27 7/9] and not the approximation 27.7 m/s.
 

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