# potential energy

• Jan 5th 2009, 06:36 PM
maeca
potential energy
this is the problem

a girl slid along an 8.0 m long frictionless slide with an inclined at 25 degrees with the horizontal. if she started from rest, what is her velocity at the bottom?

here's my question..

is the acceleration of the girl equal to 9.8m/s2sin25?

thanks
• Jan 5th 2009, 06:44 PM
vincisonfire
Yes it is. If you draw a diagram, using trigo identities you find that
$\displaystyle a = g sin(\theta)$
• Jan 6th 2009, 03:25 AM
HallsofIvy
But you don't really need to use that. Since there is no friction, you have "conservation of energy". At the top of the slide, the girl is $\displaystyle 8 sin(\theta)$ m above the base and so her potential energy, relative to the bottom of the slide, is $\displaystyle 8 mg sin(\theta)$ and her kinetic energy is 0 so her total energy is $\displaystyle 8 mg sin(\theta)$. At the bottom of the slide, her potential energy, again relative to the bottom of the slide, is, of course, 0 so here kinetic energy must be $\displaystyle (1/2)mv^2= 8 mg sin(\theta)$ or $\displaystyle v= 4\sqrt{g sin(\theta)}$. If you use acceleration to find velocity, you can use that as a check.
• Jan 6th 2009, 03:53 AM
maeca
thanks
thanks very much