(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Assume you are on a planet simillar to Earth where the acceleration due to gravity is 10 m/s^{2}. A block of mass 15 kg lies on an incline plane. The height of the incline in 9m and the width is 12m. The coefficient of kinetic friction is .5. The magnitude of the force necessary to pull the block up the incline is most nearly:

114N, 290N, 344N, 428N, 86N, 244N, 150N, 184N, 108N, 122N

2. Relevant equations

[tex]\Sigma[/tex]F_{y}=ma_{y}

[tex]\Sigma[/tex]F_{x}=ma_{x}

f_{k}=[tex]\mu[/tex]_{k}n

3. The attempt at a solution

Being that the block will be travelling at a constant speed, the sum of the forces in the x direction ought to equal zero. As for the y direction the sum of the forces must equal zero due to the normal forces and gravity cancelling eachother.

So for the forces in the y direction:

[tex]\Sigma[/tex]F_{y}=n-mgcos[tex]\theta[/tex]=0

The forces in the x direction:

[tex]\Sigma[/tex]F_{x}=mgsin[tex]\theta[/tex]-f_{k}=0

[tex]\theta[/tex]=tan^{-1}(.75)=36.87

n=mgcos[tex]\theta[/tex] = 150(cos36.87)=119.999

f_{k}=[tex]\mu[/tex]_{k}n=59.999

Which is clearly not close to any of the above options. I also tried using sin instead of cos incase I mixed up my directions but that yields me 90, which is close to one of the choices but not the correct answer. Thanks in advance.

Joe

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# Homework Help: Friction on an inclined plane

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