#1
hey i need some help for physics i have never had a worse teacher than my current physics one. heres the question:

A 2.00 kg stone is thrown vertically upward at 3.0 m/s, from a cliff that is 55.0m high. Find the kinetic energy of the stone just before it hits the ground below the cliff.

i suppose you disregard friction for that one ^

A 1000 kg car moves horizontally at a constant speed of 72 km/h. The efficiency of the car is 20%. The total resistance against the motion is 500 N. How much power is supplied from the gasoline to the engine?

Keep in mind it's grade 11 physics, probably not the hardest questions i just am stuck in a rut. thanks alot
#2
answer of the first question i think is 1108 joule. Repost if u wanna know how to do it
#3
Given:

m = 2.00kg
vi = 3.0 m/s
d = 55.0 m
a = -9.81 m/s2

d = (vi)(t) + .5(a)(t)^2

----------------------------------------------------------

55.0 m = (3.0m/s)(t) + .5(-9.81 m/s^2)(t)^2
55.0 m - (3.0 m/s)/(-4.90 m/s^2) = (t)
3.26 s = t

----------------------------------------------------------

vf = vi + a(t)
vf = 3.0 m/s + (-9.81 m/s^2)(3.26 s)
vf = -32.0 m/s

---------------------------------------------------------

KE = .5(m)(v)^2
KE = .5(2.00kg)(32.0 m/s)^2
KE = 1020 J

--------------------------------------------------------

Kinetic Energy is 1020 Joules.

I don't feel like doing the next one. Sorry. Hope this helps.
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#4
i wanna know how to do it. the answer he got isnt the a\one of the answers given.

a).58 kJ b) .73 kJ c) 1.09 kJ d) 1.36 kJ e) 1.54kJ
#5
The answer is c)1.09 kJ. Yes JOHNERREZ is right. I got my calculation a bit wrong.
#6
This is what i got for the first one. If its thrown up at 3m/s then at the same point on its return journey it will also be at 3m/s

so V^2=u^2+2as

u=3, a=9.81(gravitational acceleration) and s=55m

V=(root of) 9+1079.1

E=1/2mv^2

so E= 1/2 x 2 x 1088.1

the 1/2 and two cancel so the answer is 1088.1 joules or

1.09 KJ to (3 sigfig)
Last edited by Mask_of_Terror at Jan 14, 2007,
#7
It looked like you guys didn't take into account the fact that the stone is thrown upward from the cliff.

v = v0 + at

v = final velocity = 0 m/s
v0 = original velocity = 3 m/s
a = acceleration = -9.81 m/s^2
t = time
solve for t

x = x0 + v0t + (1/2) at^2
x = distance
x0 = original distance = 55.0 m
v0 = original velocity = 3.0 m/s
t = time = what you got from the first equation
a = acceleration = -9.81 m/s^2
solve for x

KE = 1/2 mv^2
KE = kinetic energy
m = mass = 2.00 kg
v = final velocity = v^2 = v0^2 + 2a(x-x0)
v0 = original velocity (after the stone has stopped rising after you threw it up) = 0
a = acceleration = -9.81 m/s^2
x = final position = 0 (the ground)
x0 = position = the height it got to after you threw it up (second equation)
#8
Quote by gibsonpenguin
It looked like you guys didn't take into account the fact that the stone is thrown upward from the cliff.


Umm...i did. Your way is far too longwinded. Chuck the stone up at 3m/s, due to conservation of energy, it will be at 3m/s at the same point on its way down.

I can't see why you would possibly want to calculate the time. Its not needed.
#9
Quote by Mask_of_Terror
Umm...i did. Your way is far too longwinded. Chuck the stone up at 3m/s, due to conservation of energy, it will be at 3m/s at the same point on its way down.

You posted as I was writing mine up, I didn't see it in the other two solutions.

EDIT: After seeing your edit- I solved for time so that I could find the position it reached after the stone was thrown up at 3.0 m/s.
#10
Quote by gibsonpenguin
You posted as I was writing mine up, I didn't see it in the other two solutions.


O ok, you are forgiven.
#11
Just use the formulas. It's really easy.

They are in your book if you don't know them already.

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#12
Quote by CroneR
Just use the formulas. It's really easy.

They are in your book if you don't know them already.

That was quite possibly the most unhelpful thing anybody could say, congratulations.
#13
so could anyone take a stab at the second? thanks for the first, so the fact that the stnoe is thrown up makes no difference?