(4%) Problem 7: Referring to the image to the right, the...

(4%) Problem 7: Referring to the image to the right, the mass of Block 1 is $m_1 = 3.62$ kg while the mass of Block 2 is $m_2 = 9.08$ kg. The coefficient of kinetic friction between Block 1 and the inclined surface is $\mu_k = 0.39$. The inclined surface is at an angle $\theta = 40.2^\circ$ above the horizontal. 50% Part (a) Using the variables from the problem statement together with the acceleration due to gravity, $g$, write an equation for the magnitude of the acceleration of the system. Hint: Based upon the numeric choices, Block 1 will accelerate up the incline while the suspended block, Block 2, accelerates vertically downward. $a = \frac{m_2 g - m_1 \sin(\theta) - \mu_k m_1 g \cos(\theta)}{m_1 + m_2}$

Transcript

00:01 Hello student in this problem in part a we have to find write an equation for the magnitude of the system's acceleration use the variables from the problem statement together with g for acceleration from gravity in order to write your equation right so the mass of block 1 is m1 and mass of block 2 is m2 g is the gravity acceleration and mu is the coefficient of friction here so first of all we will draw the free body diagram so here in the downward direction m2 g force will be there in the above direction tension force will be there and here in this direction tension force will be there now it is found that if the string is same the tension will be same now here m1 g force will be in the downward direction so there will be two components of force one is m1g cost theta and here in this direction m1g sine theta and normal force will be there suppose acceleration of this system is in this direction which is a so the friction force will be here in this direction this is friction force right now so suppose the acceleration of the system is a right now if we see this block 1 there is no motion in this vertical direction right so we can say that this normal force will be equal to m1g cos theta and we know that friction force is equal to mu into n so normal force is m1g cos theta so we can write friction force will be mu into m1g cost theta now we will write newton's law of motion for block 1 for block 1 we will write newton's law of motion so it is moving in the upward direction so the net force in this direction will be equal to tension force minus m1g sine theta and minus friction force which is mu m1g cost theta it should be equal to m1 into a now this is equation number one now we will write the newton's law for block two also so for block two the net force in the downward direction will be m2g minus t should be equal to m2 into a this is equation number two now we will add these two equation so what we will get we will get after adding equation number one plus equation number two so this is m2g minus m1g sine theta minus mu m1g cost theta should be equal to m1 plus m2 into a right so from here we can write expression for acceleration of the system will be m2g minus m1g sine theta minus mu cost theta divided by m1 plus m2 so this is the answer for part a we have to write the expression now in part b we have to find the value also so we will put the values of m2 and m1 and mu and theta right so the value of acceleration by after putting the values so this is 6 .18 meter per second square...

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