Finding the hange in inetic energy

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To find the change in kinetic energy of a 1609 kg car traveling at 11.9 m/s when it hits a tree, the initial kinetic energy is calculated using the formula KE = 1/2 mv^2, resulting in approximately 113,925 J. After the collision, the car comes to rest, making its final kinetic energy 0 J. The change in kinetic energy is therefore the initial kinetic energy minus the final kinetic energy, which equals 113,925 J. The discussion emphasizes the importance of calculating both the initial and final kinetic energies to determine the change. Understanding these concepts is crucial for solving similar physics problems.
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Homework Statement


A 1609 kg car has a speed of 11.9 m/s when it hits a tree. The tree doesn't move and the car comes to rest.
Find the change in kinetic energy of the car. Answer in units of J.


Homework Equations


KE=1/2mv^2



The Attempt at a Solution


Okay so like I am not completely sure where to go here...
I started out in finding the kinetic energy
KE=1/2(1609)(11.9)^2=113925.245
Any help would be super awesome.
 
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sylavel said:

Homework Statement


A 1609 kg car has a speed of 11.9 m/s when it hits a tree. The tree doesn't move and the car comes to rest.
Find the change in kinetic energy of the car. Answer in units of J.


Homework Equations


KE=1/2mv^2



The Attempt at a Solution


Okay so like I am not completely sure where to go here...
I started out in finding the kinetic energy
KE=1/2(1609)(11.9)^2=113925.245
Any help would be super awesome.

OK You found the Kinetic energy for before impact.

What is the KE after impact? What is v?

What is the difference?
 
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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