Clarification on KE and velocity direction

[MrNeuro]

In TPRH Science Workbook theres a question that says if a bomb is dropped while the plane is moving horizontally @ 300 m/s and the bomb is 400 N what is the KE of the bomb. the answer is obviously

1/2(40)(300^2)

Then they go on to give another situation where the plane is moving @ 300 m/s vertically and everything else is the same. What is the kinetic energy of the bomb now.

i got the answer right initially by just mindlessly plugging into 1/2mv^2 but during my post analysis i read that KE is a scalar hence does not care about the direction of the velocity.

Heres where i get tripped up the work energy theorem states that W=change in KE. Where W=Fcos(theta)d clearly work cares about the direction as its considered by the Cos theta value why doesn't KE care is it because the "change" in Kinetic Energy (KEf-KEi) already takes care of the "direction"?

if I'm unclear please tell me so i can further expound what I'm asking.

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[MedPR]

In TPRH Science Workbook theres a question that says if a bomb is dropped while the plane is moving horizontally @ 300 m/s and the bomb is 400 N what is the KE of the bomb. the answer is obviously

1/2(40)(300^2)

Then they go on to give another situation where the plane is moving @ 300 m/s vertically and everything else is the same. What is the kinetic energy of the bomb now.

i got the answer right initially by just mindlessly plugging into 1/2mv^2 but during my post analysis i read that KE is a scalar hence does not care about the direction of the velocity.

Heres where i get tripped up the work energy theorem states that W=change in KE. Where W=Fcos(theta)d clearly work cares about the direction as its considered by the Cos theta value why doesn't KE care is it because the "change" in Kinetic Energy (KEf-KEi) already takes care of the "direction"?

if I'm unclear please tell me so i can further expound what I'm asking.

Yes.

Work is different because work is based on displacement and force, not necessarily velocity. For instance, if push on a block with a 100N force at 60degrees to the horizontal, your work will be W=50d. If you push on a block with the same 100N force at 90 degrees to the horizontal, work will be 0.

The vector velocity, however, already accounts for the direction since velocity has magnitude and direction. In other words, you can move at 10m/s in any imagineable direction and still have the same kinetic energy.

I tried to think up a more way to say the following, but this is all I could come up with.

Work is dependent on the actual force making the object move. Kinetic energy, however, only cares about how fast the object is moving, and not about what or why is causing its movement.

 

[MT Headed]

Work does not care about direction.

The mathematical vector algebra answer is that work is the 'dot product' of the F vector and the d vector, which results in a scalar.

The MCAT answer is that work is defined as the product of the magnitude of the F vector, the magnitude of the d vector, and the cosine of the angle between them. Carefully notice that the direction of F and d don't really matter (the theta is the theta between F and d, not the theta between F and horizontal), rather what matters is how much F and d align with each other. The more they align, the bigger the resulting work will be.