Work & Velocity: Solve 100m Distance, 2 Ton Car

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SUMMARY

The discussion focuses on calculating the speed of a 2-ton car after it travels a distance of 100 meters while experiencing a frictional force of 1.2 kN. The initial speed of the car is 80 km/h (22.22 m/s). Participants emphasize using the equation of motion, specifically v² = u² + 2as, to determine the final speed after accounting for deceleration due to friction. The frictional force is derived from the car's mass and the specified friction coefficient.

PREREQUISITES
  • Understanding of basic physics concepts such as force, mass, and acceleration.
  • Familiarity with the equations of motion, particularly v² = u² + 2as.
  • Knowledge of converting units, specifically from km/h to m/s.
  • Ability to calculate frictional force based on mass and friction coefficient.
NEXT STEPS
  • Study the derivation and application of the equations of motion in physics.
  • Learn about the principles of friction and how it affects motion.
  • Explore the concept of deceleration and its calculation in real-world scenarios.
  • Investigate the impact of different friction coefficients on vehicle performance.
USEFUL FOR

Students studying physics, automotive engineers, and anyone interested in understanding vehicle dynamics and the effects of friction on motion.

herbgriffin
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Homework Statement


The driver of a car wishing to save the fuel takes his leg off the accelerator when the car is moving at a maximum speed of 80 kmph. Determine the speed of the car after traveling a distance of 100m. The mass of car is 2 tons and the force of friction is 0.6 kN/ ton.


Homework Equations


Work = Force * distance
kinetic force = 1/2 (mV^2 - mVi^2)



The Attempt at a Solution


Frictional force = 0.6kN/ton( 2 tons) = 1.2 kN
Vi= 80 kmph = 22. 22 m/s
i don't know how to start please help me ...please..
 
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use equation of motion
v^2 = u^2 + 2as
 
Why are you trying to calculate work? The problem does not ask for that. You have the friction force so can calculate the deceleration (negative acceleration). Use the formula that ashishsinghal gives.
 

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