Question

Work Power Energy

Answer 1

for those who wanna help thankyou!!

Answer 2

@ganeshie8 if u want to help again if u want only.. thank you!

Answer 3

For part \(a\) :

recall the definition of power, \(P = \dfrac{\Delta W}{\Delta t}\)

and the work energy theorem : \(\Delta W = \dfrac{1}{2}m{v_f}^2-\dfrac{1}{2}m{v_i^2}\)

Answer 4

Hello I'm back. In the problem so I'll convert MW to W right? Hmm so 1,300,000 watts

Answer 5

8.8 mins = 528 seconds

Answer 6

I got 6864x10^5 as Work

Answer 7

To calculate force at any instant, just find the velocity and use\[P = F v \Rightarrow F=\frac{P}{v}\]

Answer 8

By work energy theorem this is my set values \[6864x10^5 = (1/2)(m)(31)^2 - (1/2)(m)(5.2)^2 \]

Answer 9

is my equation right or nah?

Answer 10

I use shift solve so I got for m is 1,469,870.23

Answer 11

Looks good!

Answer 12

Okay then so moving to b.)

Answer 13

okayy so how to do this hmm

Answer 14

use the same equation from part a

Answer 15

Let the train start at time, \(t=0\), then after \(t\) seconds, below equation holds, yes ?

\[1.3*10^6 = \dfrac{\frac{1}{2}m*v^2-\frac{1}{2}m*5.2^2}{t-0}\]

Answer 16

simply plugin \(t=5.9min\) and solve \(v\)

Answer 17

Wow didn't think that you can do that can of formula. So you start with a P=W/T then you substitute the work energy theorem so you can get that formula. Genius!! Okay will calculate!!

Answer 18

I got 25.59 m/s !!

Answer 19

wolfram says the same
http://www.wolframalpha.com/input/?i=solve+1.3*10%5E6+%3D+%5Cdfrac%7B%5Cfrac%7B1%7D%7B2%7Dm*v%5E2-%5Cfrac%7B1%7D%7B2%7Dm*5.2%5E2%7D%7B5.9*60%7D%2C+m+%3D++1469870.23

Answer 20

@ParthKohli I think that formula works only if the Force is constant, looks we need to find acceleration by differentiating the velocity here

Answer 21

oh okay then

Answer 22

@ganeshie8 Yes it is, but I'm talking at a particular instant.

Answer 23

So moving to c.) hmm

Answer 24

for part c, you may "implictly" differentiate below equation both sides with respect to time and solve \(\dot{v}\)
\[1.3*10^6 = \dfrac{\frac{1}{2}m*v^2-\frac{1}{2}m*5.2^2}{t-0}\]

Answer 25

may be differentiate in below form :
\[1.3*10^6 t =\frac{1}{2}m*v^2-\frac{1}{2}m*5.2^2 \]

your goal is to find \(\dot{v}\)

Answer 26

once you have \(\dot{v}\), you can multiply that by \(m\) to get the force responsible for causing that acceleration

Answer 27

a bit confused... But wait I'll digest it

Answer 28

you know how to "implicitly" differentiate an equation, don't you ?

Answer 29

I think I forgot the concept.. Sooooo there

Answer 30

thats okay, we don't need implicit differentiation here. lets do it in the regular way

Answer 31

solve \(v\) first in below eqn :
\[1.3*10^6 = \dfrac{\frac{1}{2}m*v^2-\frac{1}{2}m*5.2^2}{t-0}\]

Answer 32

25.59 m/s is the answer there right? The b.) answer

Answer 33

just express \(v\) in terms of \(t\)

Answer 34

don't plugin \(t\) yet

Answer 35

oh okay then wait

Answer 36

take ur time

Answer 37

Hmmm I'm getting confused.. How could I solve that if I would not plug in the t?

Answer 38

just isolate it, it will be a function of \(t\)

Answer 39

Easy, its just algebra

Answer 40

\[1.3x10^6 = ((1/2)m(v^2-5.2^2)/t-0)\]

Answer 41

so I'll factor out 1/2m right then put it in the other side?

Answer 42

we have :
\[1.3*10^6 = \dfrac{\frac{1}{2}m*v^2-\frac{1}{2}m*5.2^2}{t-0}\]

\(1.3*10^6t = \frac{1}{2}m*v^2-\frac{1}{2}m*5.2^2\)

\(1.3*10^6t - \frac{1}{2}m*5.2^2=\frac{1}{2}m*v^2\)

divide 1/2m both sides :
\(\dfrac{2*1.3*10^6t}{m} - 5.2^2= v^2\)

take sqrt :
\(\sqrt{\dfrac{2*1.3*10^6t}{m} - 5.2^2}= v\)

Answer 43

plug in the value of \(m\) and simplify if you want to

Answer 44

how about t? Still will not be plug?

Answer 45

plugging \(t\) just gives the same velocity, 25.59 m/s
what good is that ?

Answer 46

two unknowns then so I have to make 2 equations too?

Answer 47

and why do you so much want to plug t = 5.9min ha

Answer 48

hmmm I'm a bit fazed sorry.. Maybe it is the heat lol carry on.. my apologies sir

Answer 49

\(\sqrt{\dfrac{2*1.3*10^6t}{m} - 5.2^2}= v\)

first, plug the value of \(m\) on left hand side and simplify

Answer 50

don't touch \(t\) ok

Answer 51

okay okay will do

Answer 52

the m has a value of 1469870.23 sooo

Answer 53

just a number, plug it in and simplify

Answer 54

after simplifying, im getting
\[\sqrt{1.76886 t-27.04}=v\]

Answer 55

this is odd I'm getting a math error wait

Answer 56

Oh okay I subtract 5.2 so that's whyyy okay I got it

Answer 57

this should be easy, just slow down, take ur time, no hurry

Answer 58

(5.2)^2 will not be touch okay got it. (2)(1.3x10^6)/(1469870.23) = 1.7689 then 5.2^2 is 27.04 okay I got at last

Answer 59

good, so now we have \(v\) as a function of \(t\) :

\[v(t) = \sqrt{1.76886 t-27.04}\tag{1}\]

Answer 60

Can you find acceleration as a function of \(t\) from above ?

Answer 61

I'll be using Kinematics?

Answer 62

How good are you at calculus ?

Answer 63

familiar with taking derivatives and integrals ?

Answer 64

I got a C in diff and a C+ for integral calc.... Yeah I suck ikr lol

Answer 65

but yeah I'm know that

Answer 66

thats good enough, so whats the relation between velocity and acceleration ?

Answer 67

dv/dt = a right?

Answer 68

Yes, we use that now.
differentiate the velocity function that we have got earluer and find the acceleration function

Answer 69

\[v(t) = \sqrt{1.76886 t-27.04}\tag{1}\]

\[a(t)=\dfrac{d}{dt}v(t) = ?\]

Answer 70

Oh my. That's right.. wow. I totally forgot that. Okay will do will do

Answer 71

once you have the acceleration function, you can plugin \(t=5.9min\) to get the acceleration at that time

Answer 72

then finally multiply that acceleration by the mass to get the force at that time

Answer 73

\[\frac{ 177 }{ 20\sqrt{177x-2704} }\]

Answer 74

a(t) =

Answer 75

right? Then plug it the 5.9 aka the 354 sec?

Answer 76

hmmm wait.. I think differentiating is wrong

Answer 77

http://www.wolframalpha.com/input/?i=%28%5Csqrt%7B1.76886+t-27.04%7D%29%27+

Answer 78

then I'll plug in 354 and I got 0.03613

Answer 79

looks good
http://www.wolframalpha.com/input/?i=%28%5Csqrt%7B1.76886+t-27.04%7D%29%27+at+t%3D5.9*60

Answer 80

multipying it to m.. I got 53106.411 N

Answer 81

thats the acceleration of train at t = 5.9 min
multiply that by the mass to get the force at t = 5.9min

Answer 82

thats it

Answer 83

okay then at last.. This problem is very tricky for me I never taught that I'll be using that kind of method.. Seems a long journey for me to fill.

Answer 84

Moving on then to letter d.)

Answer 85

so position as a function of t or nah?

Answer 86

Yes, we can still use the velocity function

whats the relation between position and velocity ?

Answer 87

you'll have to integrate velocity to get the position

Answer 88

Yes, do it..

Answer 89

integrate between t = 0 and t = 8.8min

Answer 90

http://www.wolframalpha.com/input/?i=integrate+sqrt%281.76886x-27.04%29 legit?

Answer 91

Hey it seems our velocity function has a mistake

Answer 92

Lets fix it quick

Answer 93

i think you might change the variable here: \((\frac{1}{2}mv^2)' = (\frac{1}{2}m\dot x^2)' = m \dot x \ddot x = m \dot x^2 \frac{d \dot x}{dx} = 1.3MW\)

you still have to find the integration constant but it just looks easier...i think:p

i mean for part d).

Answer 94

okay then

Answer 95

It should be :

\[v(t) = \sqrt{1.76886 t\color{red}{+}27.04}\tag{1}\]

go ahead find acceleration again, have fun :)

Answer 96

Okay the new Force is 50864.40 okay... Moving to d.)