Calculating fn on a curved ramp.

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To calculate the normal force (fn) of a car on a curved ramp, the elastic potential energy during launch is converted to gravitational energy at the peak height. The normal force can be determined using the equation fn = mgcos(ø), where ø is the angle of the ramp at that height. Understanding the forces acting on the car, particularly during its momentary stop at the peak, is crucial, as the net force is zero at that point. The discussion also highlights the importance of applying Newton's second law (F = ma) to analyze the forces along the ramp. Clarification on the ramp's curvature is necessary for accurate calculations.
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Homework Statement



A car is elastically launched up a ramp and we are required to calculate the normal force of the car at certain heights up the ramp.

Homework Equations





The Attempt at a Solution


I am able to calculate the elastic potential energy of the car during launching and i know when it reaches the targeted height it will momentarily stop and at that point all the energy is Eg. I am to calculate the vertical height and i need the ø to find fn for at that point fn is mgcosø. Any ideas how i can solve and find fn for different positions on the ramp? Maybe i am thinking about this wrong.
Thank you for your help
 
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What kind of ramp is this exactly?

EDIT: Just saw in the title that it's curve. Assuming you mean a loop, the hint is that you know what F = ma is.
 
I know when it reaches that certain height it will be at rest, so fnet is zero, so acceleration is zero. but i know at the beginning fnet is fx, fx = ma?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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