Work done against resistive forces by comparing Ek and work done

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SUMMARY

The discussion centers on calculating the work done against resistive forces for a car with a mass of 900 kg that accelerates from 0 to 26.8 m/s in 9.2 seconds, with a constant engine thrust of 3500 N. The total work done by the engine is calculated as 431,480 J, while the kinetic energy gained by the car is determined to be 110,000 J. The difference between these two values represents the work done against resistive forces, confirming the calculations presented in the discussion.

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g9WfI
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Homework Statement
A car of mass 900kg accelerates from 0 to 26.8 m/s in 9.2 s. The thrust from the engine is constant at 3500 N during this period. By comparing the kinetic energy gained by the car with the total work done by the engine, calculate the work done against resistive forces.
Relevant Equations
suvat, kinetic energy, work done
IMG_86F1E608F5D0-1.jpeg

Missed units in the photo - J

Answer is 110 000 J
 
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g9WfI said:
Homework Statement:: A car of mass 900kg accelerates from 0 to 26.8 m/s in 9.2 s. The thrust from the engine is constant at 3500 N during this period. By comparing the kinetic energy gained by the car with the total work done by the engine, calculate the work done against resistive forces.
Relevant Equations:: suvat, kinetic energy, work done

View attachment 284260
Missed units in the photo - J

Answer is 110 000 J
3500 * 123.48 = 431480, not 43300
 
Steve4Physics said:
3500 * 123.48 = 431480, not 43300
Thank you so much!
 
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