energy conservation (urgent)

energy conservation

1. potential lost = kinetic gained
potential = mgh = 0.5mgL
kinetic at start =0.5mv2

10L = v2

F = ma = deceleration * mass = frictional force = 0.2 * m
a = -0.2

v2 = u2 + 2as
in this case v2 = stop = 0, u2 = 10L
0 = 10L - 0.4s L = 0.4
0.4 s = 10 * 0.4
0.4 s = 4
s = 10 meter
remember to work in SI unit
because the answer given by the book is wrong



Potential energy of cab when wire snaps = all the energy it has to lost before comes to rest = lost to friction = force x distance
mgh = Fd
1800 * 10 * 3.7 = 4400 * d

2011-10-15 22:26:17 補充：
Oh I missed 2.

I think 天同 mistaken 63, it wasn't talking about how high it will go at the end of the flat part, it said the flat part have friction, its asking for the distance it will stop on the flat part.

I got it wrong as well, to further elaborate the question, since the answer is 10m

2011-10-15 22:27:31 補充：
it means that it will slide for 1000cm = 40cm * 25, = 25 trips on the flat surface, 2 trips return to the left edge, so 25 trips mean 24 round trips + 1 final trip, answer should be, 40cm from the left edge.

2011-10-15 22:35:57 補充：
I think for 2 it just means that length for k2 is l, and length for other strings is l / cos(theta).

63. speed of the particle when it reaches the flat part
= square-root[2gh] = squre-root[2 x 10 x 0.2] m/s = 2 m/s
where g is the acceleration due to gravity, taken as 10 m/s2

Kinetic energy of particle after passing through the flat part
= [(1/2)m(2x2) - (0.2mg).(0.4)] J = 1.2m J
[where m is the mass of the particle]

Hence, height h' reached by particle at the other end, by conservation of energy,
mgh' = 1.2m
i.e. h' = 1.2/g = 0.12 m = 12 cm

65. (d) Since all potential energy initially possessed by the lift cab has been consumed by friction as heat when the cab finally comes to a stop. Hence,
work-done by friction = loss of potential energy of the cab
(4.4 x 10^3)s = 1800g x 3.7
where s is the total distance travelled by the lift cab before coming to stop
solve for s gives s = 15 m

66. I don't think you could get a reasonable answer without knowing the natural lengths of the three strings. Are the natural lengths of the three strings the same?


2011-10-15 09:42:31 補充：
For Q63, the answer you gave may be wrong.