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Change in kinetic energy

  • Thread starter forty
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  • #1
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If an object goes from moving at 10 m/s eastward to moving at 30 m/s southward, how does its kinetic energy change?


If the object is to to change its velocity from east to south it would have stop all of its eastwards motion and then move in the southwards direction?

i have no idea if this is correct but i did the following

.5m(10)^2 + .5m(30)^2

and get 500m (mass is not given in the problem)

Any help would be greatly appreciated!
 

Answers and Replies

  • #2
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If an object goes from moving at 10 m/s eastward to moving at 30 m/s southward, how does its kinetic energy change?
This object is not moving in a one-dimension but in two-dimension. So, you can't just add/subtract them. You can use the Pythagorean theorem to find the change in velocity.

(mass is not given in the problem)
You don't need the mass. Since you are asked to find the change in kinetic and not the numerical values for the initial and final kinetic energy[tex]\Delta K=\frac{1}{2}m(v_{f}-f{i})^2[/tex]
 
  • #3
tiny-tim
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If an object goes from moving at 10 m/s eastward to moving at 30 m/s southward, how does its kinetic energy change?

If the object is to to change its velocity from east to south it would have stop all of its eastwards motion and then move in the southwards direction?
Hi forty! :smile:

No … kinetic energy is a number, not a vector.

So it doesn't obey vector addition … it obeys ordinary addition!

What's the KE before? What's the KE after? How do you subtract one from the other? :biggrin:
 
  • #4
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so are you saying its just

.5m(30)^2 - .5m(10)^2 = 400m (J)

??
 
  • #5
tiny-tim
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Yup! :smile:

(of course, the result will be in J only if the mass is in kg)

Sometimes physics really is that easy! :biggrin:
 

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