energy lost due to friction


by BunDa4Th
Tags: energy, friction, lost
BunDa4Th
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#1
Oct14-06, 08:52 PM
P: 191
A crate of mass 11.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 16.5 with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.65 m.

(b) How much mechanical energy is lost due to friction?

well, this is what i came up with

to find the energy lost i thought F - f

so to find f = u_kN

N = (mgsin16.5)

f = .400(30.62)

f = 12.25

100 - 12.25 = 87.75 is incorrect

what am i doing wrong?
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Andrew Mason
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#2
Oct14-06, 09:33 PM
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P: 6,591
Quote Quote by BunDa4Th
A crate of mass 11.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 16.5 with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.65 m.

(b) How much mechanical energy is lost due to friction?

well, this is what i came up with

to find the energy lost i thought F - f

so to find f = u_kN

N = (mgsin16.5)

f = .400(30.62)

f = 12.25

100 - 12.25 = 87.75 is incorrect

what am i doing wrong?
You are forgetting the units. You are equating force to energy. The units of energy are force x distance.

The energy lost is the work added by the pulling force less the work done by the friction force.

AM
BunDa4Th
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#3
Oct15-06, 01:48 AM
P: 191
I dont understand what you mean. The teacher decided to skip most of the stuff on this chapter and basically i am learning this on my own and from the book i got to read which i dont understand from the reading.

gunblaze
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#4
Oct15-06, 02:10 AM
P: 186

energy lost due to friction


Energy loss is simply Kinetic friction multiply by the distance travelled. To find kinetic friction, take coefficient of friction multiply by normal force acting on the block. Is the answer 23.8J? I'm not sure if i'm right also.
BunDa4Th
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#5
Oct15-06, 02:13 AM
P: 191
that answer is incorrect.

work done by gravity is -172.99
gunblaze
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#6
Oct15-06, 02:24 AM
P: 186
wat is the answer?
BunDa4Th
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#7
Oct15-06, 02:25 AM
P: 191
the book doesnt give an answer because its an even # problem, which doesnt help me at all since i cant use any other answer to figure this one out since this is the only problem in the book that ask this.

forgot to mention this is also on an online homework I have that is how I know i been getting the wrong answer.

oh yea i mention work done by gravity because that was the first part of the problem i needed to find and thought maybe i need to use that somehow.
gunblaze
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#8
Oct15-06, 02:27 AM
P: 186
then how u know that ur answer is wrong?
gunblaze
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#9
Oct15-06, 02:31 AM
P: 186
oops. I've forgot to add in the gravitational field strength. Is the answer 233.8J?
BunDa4Th
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#10
Oct15-06, 02:32 AM
P: 191
okay, that is the correct answer.

Can you explain how you get that?
gunblaze
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#11
Oct15-06, 02:37 AM
P: 186
wats the value for g?
BunDa4Th
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#12
Oct15-06, 02:39 AM
P: 191
is g = -9.8?
gunblaze
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#13
Oct15-06, 02:42 AM
P: 186
so 233.8J is right?
BunDa4Th
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#14
Oct15-06, 02:43 AM
P: 191
yea, that is the correct answer
gunblaze
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#15
Oct15-06, 02:52 AM
P: 186
Alright. Its just the same as how i do it in post #4. To find the energy lost due to friction, u need to take kinetic friction multiply by the distance travelled. The way u calculate normal force is wrong. It should be mgcos16.5 and not mgsin16.5. After calculating the kinetic friction, multiply it by the distance travelled, ie: 5.65 to solve for ur energy loss due to friction.

Dun get lost with the many values given in there. Just do what u need to do according to the formulas. Some of the values in the question are just there to mislead you. You dun necessarily have to use them all everytime. Cheers!
BunDa4Th
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#16
Oct15-06, 02:54 AM
P: 191
ahhhhh.....thanks so much. yea those other number were just there and i thought okay those number have to be there for a reason.
gunblaze
gunblaze is offline
#17
Oct15-06, 03:01 AM
P: 186
not a problem.


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