Newton's Law Problem: Calculating Acceleration on an Inclined Plane

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A block of mass m1=2.0 kg on a frictionless incline at 20 degrees is connected to a block of mass m2=1.0 kg over a pulley, prompting a calculation of acceleration. The correct formula for acceleration is derived as a=(m2g - m1gsin(angle))/(m1 + m2). Initial calculations led to confusion, with incorrect values for m1g and m2g. After clarification and using the correct formula, the acceleration was found to be -1.03 m/s² downwards. Understanding the derivation of the formula is emphasized as crucial for solving similar problems.
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Homework Statement


A block of mass m1=2.0 kg on a frictionless inclined plane of angle 20 degrees is connected by a rope over a pulley to another block of mass m2=1.0kg. What are the magnitude and direction of the acceleration of the second block?

Homework Equations


a=m2g-m1gsin(angle)/ma1+m2


The Attempt at a Solution



how would i calculate m1g and m2g in the first place?
other then that...

a=m2g-m1gsin20/2+1



Your help will beeee very appreciated seeing as I have a test on it tom!
 
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g is acceleration due to gravity = 9.81 m/s2
 
so basically i would multiply mass with 9.81
m2 = 1(9.81) = 9.81
m1= 2(9.81) = 19.62

a= -19.62-(-9.81)sin20/3
a= 3.3m/s^2 (down)

but the answer is supposed to be -1.03 m/s^2 (down)
 
Smile101 said:
so basically i would multiply mass with 9.81
m2 = 1(9.81) = 9.81
m1= 2(9.81) = 19.62

a= -19.62-(-9.81)sin20/3
a= 3.3m/s^2 (down)

but the answer is supposed to be -1.03 m/s^2 (down)


Well for one, it seems that you switched around m1g and m2g. It should be

a=\frac{9.81-(19.62)sin20}{3}


Also, do you know how to get that formula?

a=\frac{m_2g-m_1gsin\theta}{m_1+m_2}
 
rock.freak667 said:
Well for one, it seems that you switched around m1g and m2g. It should be

a=\frac{9.81-(19.62)sin20}{3}


Also, do you know how to get that formula?

a=\frac{m_2g-m_1gsin\theta}{m_1+m_2}


No i don't! But the modified version you gave me is right! I got 1.03! :) thank you soo much :cool:
 
Smile101 said:
No i don't! But the modified version you gave me is right! I got 1.03! :) thank you soo much :cool:

If the resultant force of m2 is downwards, then for the system vertically. (T=Tension)

m2a=m2g-T

therefore on the incline, m1 moves up, so that

m1a=T-m1gsin\theta


two equations where you can eliminate T by adding them. That is how to derive that formula. It's better to know how to do these kinds of of problems from first principles than to memorize a formula
 
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