How Much Work Is Needed to Alter a Crate's Velocity?

[jtc143]

Homework Statement

A 34.0 kg crate is initially moving with a velocity that has magnitude 3.70 m/s in a direction 37.0 degrees west of north. How much work must be done on the crate to change its velocity to 6.08 m/s in a direction 63.0 degrees south of east?

Homework Equations

W=1/2mvf^2-1/2mvi^2

The Attempt at a Solution

I attempted this by breaking the velocities into their components and got -2.23m/s in the initial x, 2.95m/s in the initial y, 2.76m/s in the final x, and -5.42m/s in the final y. Then I did the work-energy theorem for both x and y components and got work in x direction to be 44.96J and work in the y direction to be 351.46J. Then I did the Pythagorean theorem to get work to be 354.32J But this is wrong, can anyone help?

 

[jtc143]

Never mind I figured it out. The angles mean nothing and you just do the work energy theorem normally. However, I'm not sure WHY this is so; can anyone explain why the angles do not effect the question?

 

[PhanthomJay]

Work, like energy, is a scalar quantity, and as such, it has no direction, only magnitude, which may be a positive or negative value. In your first attempt when you broke up the velocity into its vector components, if you did the math correctly, you get the same answer as the correct answer, provided that when you get the two values of work you then add them up algebraically, not vectorially, that is, you don't use Pythagorus (there is no direction associated with work, so just add them up).