How Can I Calculate the Final Velocity of a Wagon on a Hill Using Physics?

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To calculate the final velocity of a wagon rolling down a hill, energy conservation principles can be applied, using potential and kinetic energy equations. The initial potential energy (mgh) and initial kinetic energy (1/2 mv1^2) equal the final potential energy (0, at the bottom) and final kinetic energy (1/2 mv2^2). The equations yield consistent results, confirming that energy is conserved during the motion. The calculations can also be approached using the equations of motion, where v^2 = v0^2 + 2ax, with height as the distance. Both methods lead to the same final velocity, illustrating the relationship between height, initial velocity, and gravitational acceleration.
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There's a wagon standing on a hill
The Height :0,81m
When v1 = 0m/s then v2 = 4m/s when the wagon has rolled down the slope.
What is the velocity when the startvelocity is 2m/s ...

How do I do this...

I've tried many things but I don't know the mass so how should I do?

Please help
 
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try using potential and Kinetic energy equations
 
Is this right?

Is this right then... I take 2m/s which is v1 and put in in kinetical energy equation and take out v^2= Sqrt 2*g*h = 4,45 which is the right answer but I don't seem right it must be harder than that
 
Energy is conserved;
Eintial = Efinal
Pei + Kei = Pef + Kef
mgh + 1/2mvi^2 = 1/2mvf^2

or from equations of motion:
v^2 = v0^2 + 2*a*x where x is h

both result in the same answer.
 
Last edited:
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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