potential energy and conservation of energy?

Question

can somebody help me with this problem? http://blog.360.yahoo.com/blog-CMc13Q8ycqhQA2zD4EelxBwCrdYkuhw-?cq=1
you need to go to this link because there is a picture involved.
can you post answer showing your work so I can understand better. thanks.

Answer 1

#1
Poit (A)
Ua = mgh=1000 x 9.81 x 30=294300 Joules
or Ua=294.3 kJoules
Ka= .5mv^2=0 since va=0

Point (B)
Ub=mgh=0 since h=0

Kb=Ua=.5mv^2=294300 Joules (Potential energy from point A the top was converted to Ke on the bottom point B)
Vb=sqrt(2Kb/m)=sqrt(2 x 294300/1000)=24.5 m/s

Point (C)
Ec=Kc + Uc
and Ec= Ua=Kb (law of conservation of energy)
Uc=mgh(c)=1000 x 9.81 x 10=98100 joules
Kc=Ec-Uc=294300 - 98100= 196200 Joules
Vc=sqrt(2 Kc/m)=19.8 m/s

Point (D)
Solved the same way as Point (C)
Ed=Kd +Ud
again Ed= Ua=Kb (law of conservation of energy)
Ud=mgh(d)=1000 x 9.81 x 20=196200 J
Kd=98100 Joules
Vd= 14.0 m/s

Point (E)
Ue=0
Ke=Kb=Ua law of conservation of energy
Ve=Vb

Point (F)
This one I leave up to you
You must get the same results as at point C.


#2
Now we have to consider work done against friction
Force of friction f = uN=u mg
W=f S= u mg S (S- displacement along which friction was present)
(a) v=0.5 V(point F)
V(at F)=Vc=19.8 m/s
W=Kf - .5m (.5Vf)^2 (Kinetic energy at F less kinetic energy when speed is half then that at F)
W=.5 m(1 - .25)(Vf)^2= 147015
This is work done due to friction
the since W=fS
S=W/f=W/( u m g )=147015/(.55 x 1000 x 9.81)=27.25 meter

(b)
This time W=Kf
S=Kf/(u m g)
S=36.4 m

(c) The larger the mass the more energy it will have. Energy is proportional to mass.


Problem #3
Energy stored in a compressed string
Us=.5 k x^2
k - spring constant
x -amount of compression

(a) Referring to problem #2 at point F part (a)
Us=W=.5 m(1 - .25)(Vf)^2= 147015 Joules
then x=sqrt(2 x Us/k)
x=sqrt(2 x 147015/1000)=17.2 m

(b) in this case Ef=Us
Referring to Energy computation at point F
Ef=196200 joules
then
x=sqrt(2 x 196200 /1000)=19.8m

(c) Again mass is directly proportional to energy.

Good luck.
Get back to me if you have any questions.

Answer 2

Basically there is a lot of questions all dealing with the same principle, that total energy must be conserved. If total energy is always conserved, the following statement is always true:

change in U + change in K + Work done by friction = 0

In problem 1, we have a frictionless track so the work done by friction is zero. So the previous statement becomes:

change in U + change in K = 0

U = m g h and K =1/2 m v^2, and m and g are always constant while h and v change. We can rewrite the conservation of energy as:

m g (change in h) + 1/2 m (change in v^2) = 0

We want to find K, U, and v at 6 different points. I will find them at point A and B using the information provided. At point A velocity and K are both zero (starting at rest), U = m g h = 1000 kg *9.8 m/s^2 * 30 m = 294000 J = 294 kJ.

At point B, h is zero so all the potential energy we had at point A is now kinetic energy, K = 294 kJ. K =1/2 m v^2 so v = sqr(2 K/m) = 24 m/s.

The same process can be done for every other point.

In problem two, we can assume change in potential energy will be zero, (no change in height) so our conservation relationship is:

change in K + work done by friction = 0

Work done by friction is given by: W = F d where F is the force of friction. Solve v as a function of d.

1/2 m v^2 = mu m g d

v^2 = 2 mu g d

The whole problem is just manipulation of this relationship.

In problem 3, you have a spring. We can again assume work done by friciton is zero (frictionless surface). So,

change in U + change in K = 0

U for a spring = 1/2 k x^2, where x is your distance from the equilibrium position. This problem basically becomes exactly the same as problem 1.