the diagram shows a skateboarder starting a run from Position P (2.5m) at 2.6 m/s. The mass of the skateboarder system (boarde + board) = 74.5kg. Assume that frictional forces can be ignored.

Question

a)calculate the speed of the skateboarder at position a(1.5 m)

total Emechanical = Ek + Eg
=1/2mv^2 + mgh
=1/2(74.)(2.6)+(74.5)(9.8)(1.5)
=1346.96J

at a:
1/2mv^2 +mgh = 1346.96J (1.3X10^3)
37.25v^2 + (74.5)(9.8)(1.5) = 1.3 X10^3
v^2=5.499
v=2.34 m/s

b)calcuate the speed at position B(0m)
37.25 v^2 +0=1.3 X10^3
v^2=34.899
v=5.9m/s

c)Calculate the final height the skateboarder will reach up along the slope after passing position B (0m)

okay, so for this do I just solve for "h" using the 1/2mv^2 +mgh = 1346.96J (1.3X10^3) and subbing in the speed calculated in b?

Or do I use this formula:
v^2/2g = h

I've done both and they both give me different answers. Not sure which one to use then. Both formulas were taught in the same lesson.

Thanks!

Answer 1

okay, so for this do I just solve for "h" using the 1/2mv^2 +mgh = 1346.96J (1.3X10^3) and subbing in the speed calculated in b?

Or do I use this formula:
v^2/2g = h
---------------------------

Assuming that your 1346.96 J is correct
1/2mv^2 +mgh = 1346.96J
is the right idea because:

v^2/2g = h is really
v^2/2 = gh
or (1/2)m v^2 = gh
where all the kinetic energy at the bottom is turned into potential energy when the m stops at the top of something
and it comes from conservation of energy:
(1/2) m V1^2 + g H1 = (1/2) m V2^2 + g H2 = total energy
note that if H1 is zero and V2 is zero, you get the simple equation.

Answer 2

Note by the way that V2 = 0 if the person is up as high as possible.

Answer 3

a constant net force of 1500 N gives a troy rocket an acceleration of 2.5m/s squares.what is the mass of the rocket?

Answer 4

Force