Lab help, can't get the same result as interactive physics

In summary, a 15 kg block is pushed back a distance of 0.75 m against a spring with a spring constant of 150 N/m. This is done on an incline of 25 degrees with a coefficient of kinetic friction of 0.4. Additionally, there is a string pulling with a tension of 20 N at an angle of 10 degrees above the incline. The question asks for the distance the block moves when released. Using the equations for nonconservative work and energy, the final displacement is calculated to be approximately 0.792 m.
  • #1
Reaga
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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.
 
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  • #2
Welcome to Physics Forums.

The potential energy stored in a spring is +(1/2) k x2 (no minus sign). The rest of your work seems okay.
 

FAQ: Lab help, can't get the same result as interactive physics

Why am I not getting the same result in my lab as in interactive physics?

There could be several reasons for this. One possibility is that there may be errors in your experimental setup or data collection. Another possibility is that the simulation in interactive physics is simplified and may not accurately represent real-world conditions. It is important to carefully review your procedures and data to identify any discrepancies.

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To ensure accuracy in your lab results, it is important to carefully follow the experimental procedures and take precise measurements. It is also helpful to repeat the experiment multiple times to reduce the impact of random errors. Additionally, comparing your results with those of others or with theoretical predictions can help identify any potential sources of error.

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To improve the validity of your lab experiment, it is important to carefully control all variables and minimize sources of error. This can include using precise measuring tools, conducting multiple trials, and ensuring consistent conditions throughout the experiment. It is also important to consider any potential biases and make efforts to mitigate them.

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