Its been a year now i'm backhelp ? =D

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SUMMARY

The discussion centers on calculating the driving force required for a racing car with a mass of 1500 kg, accelerating at 5.0 m/s², while accounting for lift and ground effects. The net force equation Fnet = ma is applicable, but the impact of friction must also be considered. The frictional force is determined by the normal force multiplied by the coefficient of kinetic friction, which is given as uk = 1.0. The solution requires calculating the normal force based on the car's weight and adjusting the driving force accordingly.

PREREQUISITES
  • Understanding of Newton's Second Law (Fnet = ma)
  • Knowledge of forces acting on objects (lift, weight, and ground effects)
  • Familiarity with friction calculations (Friction = Normal Force x Coefficient of kinetic friction)
  • Basic physics concepts related to acceleration and mass
NEXT STEPS
  • Calculate the normal force for a 1500 kg car under the influence of gravity
  • Determine the total frictional force using the coefficient of kinetic friction
  • Apply the net force equation to find the required driving force
  • Explore advanced topics in vehicle dynamics and aerodynamics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the forces acting on vehicles during acceleration.

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its been a year now I'm back . help please ? =D

hey guys it's been a while i haven't been on in a long time since last year when i last had physics 10 the introductory course and now I'm back just looking for guidance how to do this question ( helping friend :!) anyways
here it is:

A racing car has a mass of 1500kg, is accelerating at 5.0m/s^2, is experiencing a lift force of 600 N up (due to its streamlined shape) and ground effects of 1000 N down (due to air dams and spoilers). Find the driving force needed to keep the car going given that uk = 1.0 for the car


since it's been so long since I've last done physics the only equation that comes to mind is Fnet = ma but I'm sure that's wrong lol sooo can someone help me out? thanks !
 
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Welcome back.

The equations you have written holds true for any situation. But in this case you may forget that whatever resultant force downward the car provides, will be provided equally in the opposite direction by the ground. So we know there will be NO net force UP or DOWN.

Can you work out the FORCE that is needed to accelerate the car by this much (neglecting friction)?

Now we have to consider friction. There will be a force of friction opposing this force you have worked out now. So we need to increase the force, by adding the magnitude of friction to the previous answer.

Friction = Normal Force x Coefficient of kinetic friction

What is the normal force? (Hint: don't forget about the objects WEIGHT).

Hope this helps,
Sam
 

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