# Work/energy

• May 9th 2008, 10:26 AM
jason03
Work/energy
heres the problem

heres my work

http://i14.photobucket.com/albums/a3...istx/faaaa.jpg

The speed is supposed to be 5.18 but im not getting that...

any ideas?
• May 9th 2008, 12:18 PM
ok first i am going to convert all the units given in tat question to SI units (i was taught in SI units)

mass of the collar, m = 0.17 kg (1 oz = 0.0283495231 kg)
spring constant, k =875.83 N/m (1 lb/in = 175.166 N/m)
distance spring compressed, x = 0.0381 m (1 inch = 0.0254 m)

Now total elastic potential energy,EPE stored in the spring before release is given by the formula,
$\displaystyle EPE = \frac{1}{2}kx^2$
$\displaystyle EPE = \frac{1}{2}(875.83)(0.0381)^2$
$\displaystyle EPE = 0.635681793J$

The collar is then released. At point B, the collar has already traveled 10 inches (0.254 m) vertically upwards.

Hence gravitational potential energy,GPE gained by the collar is:
$\displaystyle GPE = mgh$
$\displaystyle GPE = (0.17)(9.81)(0.254)$
$\displaystyle GPE = 0.4235958J$

using principle of conservation of energy the total energy of the system is conserved:

Initial energy = final energy
EPE = Kinetic energy + GPE
0.635681793J = KE + 0.4235958J
KE = 0.21085993 J

from the formula of kinetic energy, KE:
$\displaystyle \frac{1}{2}mv^2 = 0.21085993$
$\displaystyle v^2 = \frac{2(0.21085993)}{0.17}$
$\displaystyle v = 1.579597838 ms^{-1} = 5.1824 fts^{-1} \longrightarrow (1 ms^{-1} = 3.2808399fts^{-1})$

well as for the force part take note that:

elastic potential energy, EPE = $\displaystyle \frac{1}{2}kx^2 = \frac{1}{2}Fx$
• May 9th 2008, 02:13 PM
jason03
Thanks!....that really helped.....i realized one of my big problems was converting the spring constant which is what I forgot......I like working in metric better as well.....U.S. Customary is a pain.....

What is the strategy for finding the force the collar is putting on the rod?
• May 9th 2008, 09:47 PM
the motion of the collar now is in a circle, hence we can consider centrifugal and centripetal force acting on the collar.
$\displaystyle F = ma$
$\displaystyle F = mr\omega ^2$
$\displaystyle F = mr{(\frac{v}{r})}^2 \longrightarrow v = r\omega$
$\displaystyle F = m(\frac{v^2}{r})\longrightarrow (1)$

well you have already calculated the linear speed,v (do not confuse with angular speed $\displaystyle \omega$) of the collar previously also r and m are given in the question. Hence just substitute those values into equation (1) and solve for F remember to state the direction of the force on the collar as well which in this case at point B is inwards (towards the left/ towarsd the center of the circle)
• May 10th 2008, 02:39 AM
topsquark
Quote:

elastic potential energy, EPE = $\displaystyle \frac{1}{2}kx^2 = \frac{1}{2}Fx$
Be VERY careful here. $\displaystyle \vec{F} = -k \vec{x}$. If you aren't careful your spring potential energy will be the negative of what you are looking for!