Question

Can anyone help me with Work and Energy? There's a 2kg mass moving on a surface which in the end there's a spin. The distance between A-B is 5 meters and ONLY between A-B there's friction (meaning under the spin there's no friction) The initial velocity of the mass is 6.261m/s the coefficient of friction is 0.2.Find constriction of the spring when the box hits it knowing the K of the spring is 36.4 N/m

Answer 1

Attached herewith the question

Answer 2

I thought it should be W=KX^2- mv^2*0.5

Answer 3

Wf=Eel-Ek

Answer 4

Am I wrong, or the book is wrong?

Answer 5

I got 1.7974 m, is it correct?

Answer 6

@Abhisar

Answer 7

@Abhisar What do you think

Answer 8

Thanks!

Answer 9

an you calculate the final velocity with which the box collided with the spring?

Answer 10

*Can

Answer 11

6.261 m/s

Answer 12

@pooja195

Answer 13

After that use \(\sf \frac{1}{2}mv^2=\frac{1}{2}kx^2\)

Answer 14

Getting it?

Answer 15

the final? not the initial?

Answer 16

Yes, use final velocity......

Answer 17

How can I calculate it? sorry I'm a bit confused

Answer 18

the final velocity

Answer 19

and why not the way I've calculated it?

Answer 20

You can use \(v^2=u^2+2as\)

take u=6.261
s=5m
a=0.2*9.8

Answer 21

a is the coefficient of friction? I've never seen this equation

Answer 22

I have used conservation of energy, the kinetic energy of box after collision gets stored as the potential energy of spring.

Answer 23

oh kinematics right my bad lol

Answer 24

You're incredible! Thanks allot!

Answer 25

Truly brilliant

Answer 26

So does it make sense that the final velocity would be much smaller?

Answer 27

I got 7.668 m/s

Answer 28

a is acceleration, I have calculated acceleration by multiplying coffecient of friction by acceleration due to gravity.

Yes, final velocity will be smaller since friction is acting.

Answer 29

wait, but I got the same answer 1.797 m

Answer 30

The computer says I'm wrong

Answer 31

nooooooooooooooooooooooo

Answer 32

@Abhisar I got the same answer as before, by the way I've calculated it from the beginning

Answer 33

oh, i didn't check ur work...

Answer 34

are u sure, u hav calculated things correctly?

Answer 35

yeah it's the same result 1.797s m/s

Answer 36

I should join the circus instead lol

Answer 37

premed is crazy

Answer 38

but thanks! I'll ask my TA on Sunday morning

Answer 39

thank you so much!

Answer 40

np, but lemme recheck it..

Answer 41

darn

Answer 42

Calculation of final velocity

\(\sf v^2 = u^2 - 2\times 0.2 \times 10 \times 5 \\ \therefore v \approx 4.3\)

Answer 43

Now,

\(\sf \frac{1}{2}\times 2\times {4.3}^2=\frac{1}{2}\times 36.4 \times x^2\)

Calculate the final value of x

Answer 44

I got a different answer this time....

Answer 45

Really?

Answer 46

@Abhisar What did you do differently this time?

Answer 47

I have written my steps above....^^^^

Answer 48

I got 1.027m

Answer 49

did u get the same?

Answer 50

Yes, I got approximately the same...

Answer 51

but why did u multiply by 10?

Answer 52

v2=u2−2×0.2×10×5∴v≈4.3

Answer 53

v2=u2−2×0.2×10×5

Answer 54

I'm trying to copy the equation you did to get the final velocity

Answer 55

@Abhisar

Answer 56

I used 10 for ease of calculation, you can take it 9.8 if you wish.

Force of friction = Mass X coff. frictin X acceleration due to gravity

acceleration due to friction = coff. frictin X acceleration due to gravity

Got it?

Answer 57

*frictin = friction.

Answer 58

you're brilliant! Thanks allot!

Answer 59

Welcome ^_^

Answer 60

and not really, I am just ok ;)

Answer 61

and then you've multiplied it by 5 as well

Answer 62

because of the distance, right?

Answer 63

lol

Answer 64

Yes :)

Answer 65

I wish I could help you as much as you've helped me

Answer 66

^_^

Answer 67

thank you thank you thank you

Answer 68

Weeeeee ! \(\huge \ddot \smile\)