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A 2000 kg is initially traveling with a speed of 20 m/s. The driver applies the break and the truck slows to 10 m/s. How much work was done by the frictional force applied from the breaks?(A) –10,000 J(B) –5,000 J(C) 5,000 J(D) 10,000 J

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Chapter 13

Practice Test 3

Section 1

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Hi in the given problem mass of the object is M is equal to 2000 kilogram. It's initially speed is given to be v. I. is equal to 20 m/s, and the final speed is reduced to do by applying bricks that is reduced to and meter per second. Now we have to find work done by the frictional force. So using work energy theorem that work done by the frictional force will be given by change in kinetic energy knees. Final kinetic energy minus initial kinetic energy means this is half and we every square minus half and the eyes square. So plugging in or loan values this we have done is equal to we can take this half M. As a common out for em. This is 2000 kg for we have this is 10 m per second square minus 20 m per second. The square. So it comes out to be 1000 bracket 100 -400. So this is finally 1000 multiplied by minus 300 Jews. So we can say this organism minus which can be represented as minus three into 10, part five jews minus three luxury, or minus 3 to 10 days. Part five jews, which is the answer for this problem, but none of the option is correct here in this problem. Thank you.

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