How Do You Calculate the Velocity of a Mine Car Hoisted Up an Incline?

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To calculate the velocity of a 400kg mine car hoisted up an incline at t=2 seconds, the force in the cable is given as F=3200t^2 N. The angle of the incline is determined to be 28.0724 degrees, and the initial velocity is 2 m/s. The net force equation is established as E F(x)=F-mgsin(theta)=ma, leading to an acceleration of a=8t^2 - 4.616. After integrating the acceleration, the velocity equation is v=8/3 t^3 - 4.616t + 2, resulting in a calculated velocity of 14.10 m/s at t=2. The solution appears to be correct based on the calculations provided.
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Homework Statement


A 400kg mine car is hoisted up the incline (8,15,17 Triangle which i calculated the angle to be 28.0724 degrees)using a cable and motor. For a short time, the force in the cable is F=3200t^2 N, where t is in seconds. If the car has an initial velocity v=2 m/s @ t=0, determine its velocity @ t=2.


Homework Equations



F=ma

The Attempt at a Solution



Since there is no friction i believe that the Y forces don't matter nore does the Normal.
So I sumed the forces in the x direction which is parallel to the slope of the hill.

E F(x)=F-mgsin(theta)=ma i got a=8t^2 - 4.616...

i then integrated a to get v=8/3 t^3 - 4.616t+c c= 2. pluged in t=2 and got v=14.10 m/s


is this correct
 
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looks correct...
 
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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