Kinetic energy and Potential energy

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When a vehicle weighing 2.8 tons slows from 100 km/h to 50 km/h, it loses kinetic energy, which can be converted into gravitational potential energy. The kinetic energy lost can be calculated using the formula KE = 0.5 * m * v^2, where m is mass and v is velocity. This stored energy can then be used to determine how high the vehicle could theoretically climb, using the gravitational potential energy formula PE = m * g * h. By deriving the relationship between the initial and final velocities, one can calculate the height the vehicle can ascend. The discussion concludes with the successful application of these equations to solve the problem.
omni
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Vehicle with a weight 2.8 ton Slowing from 100 km/h to 50Kkm/h how much energy can was generate from this Slowing.
(if we Ignore form Friction )
and to which Height this energy was can Climb up the Vehicle Up the hill

thanks.
 
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Consider the equation for kinetic energy. If the velocity is reduced, the kinetic energy is reduced. This energy must go somewhere. Assume it is stored. Now think about the formula for gravitational potential energy. Could this stored energy be transferred to increasing the elevation of the car? Can you derive an equation of the increase in elevation of the car as a function of the initial and final velocities?
 
omni said:
Vehicle with a weight 2.8 ton Slowing from 100 km/h to 50Kkm/h how much energy can was generate from this Slowing.
(if we Ignore form Friction )
and to which Height this energy was can Climb up the Vehicle Up the hill

thanks.

Do you know the equation for the Total Energy of an object in terms of its Kinetic Energy and Potential Energy? That's the equation that you use to figure out this question.
 
yes thanks to both of you i solve it. :)
 
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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