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Re: CONSERVATION OF ENERGY



At 18:10 7/15/97 EDT, you wrote:
... Please show what one of your
best students should be able to do in about 30 minutes. A pocket
calculator is allowed.

I am my best - and only ( perennial) student.
Moreover, I shall not suffer unduly if made to lose face -
a concern for folks from particular heritages...
I hope this can serve some educational purpose?

*********************************************************************
An iron cube of mass m=1 kg slides horizontally over a long iron plate
which remains at rest. The mass of the plate is 99 kg. The specific heat
capacity of iron is 470 J/(Kg*K), the coefficient of kinetic friction is
0.57. The initial kinetic energy of the cube is 50 J. Friction brings the
cube to rest before its base reaches the plate's boundary. Everything is
in the vacuum.

1) What is the initial speed of the cube? How far will it slide?
(hint: g=9.80 m/s^2)


KE = 1/2 m.v^2 = 50J v^2 = 50 .2 / 1 = 100 m^2/s^2 v = 10 m/s

Frictional resistance = .57 x 9.8 newtons

a = f/m = 5.586/1 m/sec^2
v^2 = u^2 + 2as so 100 = 2 x 5.586 x s s = 8.95 m

2) Calculate the change of temperature of iron (both pieces treated as
one object). What assumptions are you making to find the answer?


dt x 100 x 470 = 50
dt = 0.001 degC

Assume both pieces stabilize at same temp., and radiative heat loss is
sensibly constant.

3) How much heat is generated? How much work is done? Explain what happens
to the initial kinetic energy of the iron cube.

Heat - 50 joules
Work 0 joules
The initial KE is modeled as all converted to heat.

*********************************************************************
Ludwik Kowalski
P.S.
Are words "generated" and "done" acceptable? What words would be better?
Keep it simple, please.



I don't believe my student persona is meant to answer these final
questions so I will abstain.
Regards
brian whatcott <inet@intellisys.net>
Altus OK