Problem 8.82 a previous 13 of 13 | return to assignment Problem 8.82 Example 8-1
ID: 1770388 • Letter: P
Question
Problem 8.82 a previous 13 of 13 | return to assignment Problem 8.82 Example 8-17 depicts the following scenario. A block of mass m. 2.40 kg is connected to a second block of mass m 1.80 kg as shown in the figure. When the blocks are released from rest, they move through a distance d 0.500m, at which point m2 hits the floor. If the coefficient of kinetic friction between m and the horizontal surface is 0.450, the speed of the blocks just before m2 lands is 1.30 m/s. Now we will consider a slightly different related scenario to Example 8-17 Part A Suppose we would like the landing speed of block 2 to be increased to 1.40 m/s. Should the coefficient of kinetic friction between block 1 and the tabletop be increased or decreased? it should be increased O it should be decreased Submit My Answers Give Up Correct Part B Find the required coefficient of kinetic friction for a landing speed of 1.40 Submit My Answers Give UpExplanation / Answer
a)we can write for m2 and m1
m2g - T = m2a (1)
T - u m1g = m1a (2)
solving both simultaneously and solving for a we get
a = (m2g - uk m1g)/(m1 + m2)
a = (1.8 x 9.81 - 0.45 x 2.4 x 9.81)/(2.4 + 1.8) = 1.68 m/s^2
We know that,
v^2 = u^2 + 2 a S
v = sqrt (2 a D) = sqrt (2 x 1.68 x 0.5) = 1.3 m/s
Now the distance is same, but v = 1.4 m/s
1.4^2 = 2 a D
a = 1.4^2/2 x 0.5 = 1.96 m/s^2
Again going back to the ecpression of a
1.96 = a = (m2g - uk m1g)/(m1 + m2)
1.96 (2.4 + 1.8) = (1.8 x 9.81 - uk x 2.4 x 9.81)
8.23 = 17.66 - 23.54 uk
23.54 uk = 9.43 => uk = 0.4
Hence, uk = 0.4