- #1

- 2

- 0

## Homework Statement

A 15 kg block is against, but not attached to, a spring w/ k=150 N/m and pushed back a distance of .75 m. All of this is on an incline of 25 degrees, w/ coefficient of kinetic friction equal to .4. To top it off, there's a string pulling w/ a Tension of 20 N at an angle of 10 degrees above the incline. How far does the block move when released?

## Homework Equations

Wnc=Kf-Ki+GPf-GPi+Sf-Si

Wnc= nonconservative work

K=kinetic energy

GP=Universal Gravitational Potentiential Energy=mgh

S=Spring Work=-1/2(k)(x^2)

Fk=FN*mk

Fk=friction force

FN=normal force

mk=.4

## The Attempt at a Solution

Well, first I found the x and y gravitational forces, w/ the positive x-axis pointing up the incline. That's (147sin(25),147cos(25))=(62.125,133.227)N

Then I found the same components for tension: (20cos(10),20sin(10))=(19.696,3.473)N

FN=Fwy-Ty (the problem says nothing about the block leaving the incline, and in face he told us to put a frictionless block above it in IP to prevent this). So FN=133.227-3.473=129.754 N

Fk=.4FN=51.9016 N

WT=20cos(10)d=19.696d

WFk=51.9016(cos180)(d)=-51.9016d

Now I start using energy equations (initial and final position have its velocity at 0, so K is ignored)

(15)(9.8)(dSin25)-15(9.8)(0)+0-(-1/2)(150)(.75^2)=19.696d-51.9016d

62.125d+42.1875=-32.2056d (I was already suspicious here since the answer would clearly be negative)

42.1875=-94.3306d

ignoring a negative answer, I'd get d=.447 m

IP (through Vector addition) gives me a displacement of .792m

Now I didn't do the IP myself, a partner did, but I need to know if I did anything wrong here.

Last edited: