Work done against resistive forces by comparing Ek and work done

AI Thread Summary
The discussion centers on calculating the work done against resistive forces for a car accelerating from 0 to 26.8 m/s with a constant engine thrust of 3500 N. The kinetic energy gained by the car is compared to the total work done by the engine, leading to a calculated work done against resistive forces of 110,000 J. There is a correction regarding the calculation of work done, emphasizing the importance of accurate unit representation. The final answer confirms the work done against resistive forces as 110,000 J.
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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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