Can any of you guys help to understand kinetic energy?

AI Thread Summary
Kinetic energy conservation can be applied to a system of two blocks connected by a spring, both given an initial horizontal velocity at a 45-degree angle. The discussion clarifies that while energy is a scalar and does not require splitting into components, the force responsible for the initial velocity should be considered to determine spring elongation. The correct approach involves equating the initial kinetic energy of the blocks to the potential energy stored in the spring at maximum elongation. Confusion arises regarding the angle of velocity since the blocks are constrained to horizontal motion. Ultimately, the calculation should focus on the horizontal component of the velocity for accurate results.
suryanarayan
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can any of you guys help to understand kinetic energy?

when we apply kinetic energy conservation...should we split the velocity into two components?
eg)))))
two blocks of mass m are connected by a spring and are both given a velocity v at an angle of 45 degree with the horizontal(they lie on the horizontal)
..assuming no friction is present...
what will be the maximum elongation of the spring??


i used by equating 1/2 k x2 with the initial ke of the two masses along the horizontal i.e.vcos45?is it correct?pls help me(no vertical motion)
 
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Well energy is a scalar so it it's meaningless to split the KE into two components.

You should however find the component of the force that gave the initial velocity in order to find the elongation.
 


can you pls tell me how to to solve it then...or did i do it correctly?
 
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I'm confused by the problem. How can the velocity be at a 45 degree angle with the horizontal if the masses must move only on the horizontal?
 


wotanub said:
I'm confused by the problem. How can the velocity be at a 45 degree angle with the horizontal if the masses must move only on the horizontal?

they are both given a horizontal velocity v...its given like that in the question...dont know why
 


In that case, I'd work it as you did.

2 \frac{1}{2} m (vcos(45°))^{2} = \frac{1}{2}k(Δx)^{2}
 
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