# Two blocks with masses 4.00 kg and 8.00 kg are con...

Two blocks with masses 4.00 kg and 8.00 kg are connected by astring and slide down a 30.0 degree inclined plane. The coefficientof kinetic friction between the 4.00-kg block and the plane is0.25; that between the 8.00-kg block and the plane is 0.35.
a) Calculate the acceleration of each block
b) Calculate the tension in the string
c) What happens if the positions of the block are reversed, sothat the 4.00-kg block is above the 8.00-kg block? My professor told me the answers were
a) 2.21 m/s2
b) 2.26 N
c) alower = 1.93 m/s2, aupper =2.78 m/s2

how would i arrive at these solutions?

1. First note that you must establish a reference frame (i.e. x-yplane). For convenience, it is genearlly assumed that thex-axis is parallel to the slope of the surface and the y-axis isperpendicular to the slope. Furthermore, the positivex-direction is downslope.
Given the above assumptions, now determine the sum of theforces for each block.
m1=4kg mass of block 1
m2=8kg mass of block 2
u1=0.25 coefficient of friction for block 1
u2=0.35 coefficient of friction for block 2
alpha = 30 deg --> slope of ramp.
g = acceleration due to gravity = 9.8m/s2
sum of forces on block 1:

Note that the coefficient of friction is applied to the forcenormal to the surface
sumF1=m1*g*sin(alpha) - u1*m1*g*cos(alpha) = 11.1129newtons
where m1*g*sin(alpha) is the force in the positive x-directionon block 1 due to its weight and m1*g*cos(alpha) is the normalforce, perpendicular to the slope, due to acceleration ofgravity.
Given the sum of forces for F1, it's acceleration can becalcuated as:
sumF1 = m1*a1 or a1 = sumF1/m1 = (11.1129)/(4) = 2.778m/s2
Similarly for block 2,
sumF2 = m2*g*sin(alpha) - u2*m2*g*cos(alpha) = 15.436newtons
and
sumF2 = m2*a2 or a2 = sumF2/m2 = (15.436)/(8) = 1.925m/s2
Now one must oberseve that this is not possible, that is, ifblock 1 is accelerating faster than block 2 than block 1 willinfluence block 2. This occurs through the string that isconnected between them. Note (very important) that the stringis not elastic and therefore cannot store energy. Given thisstatement, then the two blocks must move as if they were onemass. Then the total sum of forces for the two blocks combined is as follows:
sumFt = m1*g*sin(alpha) + m2*g*sin(alpha) - u1*m1*g*cos(alpha)- u2*m2*g cos(alpha) = 26.549 newtons
and the accleration of the blocks together is:
sumFt = (m1+m2)*at or at = sumFt/(4+8) = 2.2124m/s2
Now, how to figure out the tension in the string. Theeffect of the string (based on our original calculation of a1 anda2) is to apply a force to block 1 and slow it down and apply anequal but opposite force on block 2 and speed it up. So, wejust calculated the accleration for the two blocks moving together,at = 2.2124. Then we can determine the net force on eachblock as follows:
newsumF1 = m1*at = (4)(2.2124) = 8.849 newtons.
but earlier, we calcuated the net force on block 1 as 11.1129newtons. Then the difference must be the force that thestring applies to the block 1 or:
Ftension = 11.1129 - 8.849 newtons = 2.263 newtons (pointingtoward block 2)
Remember what we said about the string applying a force toboth blocks (equal in magnitude but opposite in direction)? Then the same force must be added to block 2. Lets check:
newsumF2 = m2*at = (8)(2.2124) = 17.6992 newtons, and ouroriginal calculation for sum of forces was 15.436 newtons. Therefore the string tension is:
Ftension = 17.6992 - 15.436 = 2.263 newtons (pointing downslope).
The two calcuations for tension agree!!
Now, if the blocks were reversed, what would happen? Based on our original calculations, the smaller block has higheracceleration than the larger block, therefore it would slidedownslope faster until it hit the larger block. So initialaccelerations would be a1=2.778 m/s2 for small block anda2=1.9295 m/s2 for large block until the two blocks cametogether, then the two together would move with acceleration ofat=2.212 m/s2
Hope that helps! Note that the most important concept ofthese types of problems is to determine the sum of forces andkeeping track of appropriate sign (i.e. direction of force). You should always sketch the system and forces!
Regards,
Frank