Question

I need help writing the energy conservation equation for the problem.

Answer 1

A m= 10 kg block is released from a height of h= 3 m. It travels down a frictionless circular shaped ramp to the bottom where it encounters a rough surface. It travels over this rough surface for 6 m followed by 3 m more on a frictionless surface. At this point the block compresses a spring by 0.3 m before stopping. The spring constant of the spring is 2250 N/m. Find the coefficient of kinetic friction for the rough surface

Answer 2

marry me!! :DDD

Answer 3

so far I have something like this
\[W_{NC}+U_{i}+K_{i}=U_{f}+K_{f}\]
\[(\mu*6m)+0+(10kg*9.8m/s^2)(3m)=1/2(2250N/m)(.3m)+K_{f}\]

Answer 4

I don't think this is right

Answer 5

OK

Answer 6

initially we only have gravitationaa potential energy

Answer 7

given by: Ei = m*g*h = 10kg * 9.81m/s^2 * 3m = 294.3J

Answer 8

everything right so far?

Answer 9

yes I follow

Answer 10

now the block travels down a FRICTIONLESS circular ramp

Answer 11

this means that there is no energy dissipation, onle convertion of potential energy in kinect energy

Answer 12

then it encounters a rough surface and move 6m on it

Answer 13

here we have energy dissipation in form of heat

Answer 14

how can you measure the energy dissipation in the interval?

Answer 15

it would be the friction right?
\[-\mu*6m\]

Answer 16

you forget the gravity in this equation

Answer 17

the dissipation of energy will be -mu*m*g*d = -mu*10*9.81*6

Answer 18

dissipation = -588.6*mu

Answer 19

ok that makes sense

Answer 20

after this, it travels a frictionless surface, no energy is dissipated

Answer 21

now, it compresses the spring till it stops

Answer 22

the deformation is 0.6 m

Answer 23

the energy stored in the spring for this deformation is (k*x^2)/2

Answer 24

.6? how?

Answer 25

sorry 0.3 m

Answer 26

stored = (2250*0.3^2)/2 = 101.25J

Answer 27

ok =)
\[\frac{2250*.3}{2}=337.5\]

Answer 28

I forgot to square it

Answer 29

yes

Answer 30

now make the balance of energy

Answer 31

the energy is conserved if we consider a system as a whole

Answer 32

\[294N=101.25J-588\mu\]

Answer 33

Ei = 294.3J
dissipation = 588.6*mu
stored = 101.25J

Answer 34

Ei = dissipation + stored

Answer 35

294.3 = 588.6*mu + 101.25

Answer 36

mu = 0.327982

Answer 37

...I don't know why I made that negative
.33 =D yes! thank you

Answer 38

No problem ;)