Dynamics pythagorean confusion

[Superhawkkodaka]

Homework Statement

In position 1, the 0.25kg block is held against the spring, compressing it by 150mm. the block then is released, and the spring fires it up the cylindrical surface. Neglecting friction, find the contact force exerted on the block by the surface in position 2.[/B]

1/2kx^2 - mgr(1-cos30) = 1/2mv^2>>

just wondering.. how come the angle for r is (1-cos30)? i don't think pythahorean theorem applies in this problem since it's a circular path.. how come it's 1-cos30?

also why it's negative for potential energy? >> -mgr(1-cos30)

 

[Chestermiller]

r(1-cos30) is the vertical distance that the block has risen. Draw a horizontal line from point 2 toward the left, intersecting the vertical line, and you will see your right triangle.

Chet

 

[CWatters]

also why it's negative for potential energy? >> -mgr(1-cos30)

It's not. Start with conservation of energy..

Initial Energy = Final energy

Expand that to..

Initial Energy (in the spring) = Energy converted to PE + Remaining KE at the top

Then rearrange it to give..

Initial Energy (in the spring) - Energy converted to PE = Remaining KE at the top

or

1/2kx^2 - mgr(1-cos30) = 1/2mv^2