Question

Chapter 04, Problem 088 GO Two blocks are sliding to the right across a horizontal surface, as the drawing shows. Block 1 Block 2 m = 3,3 kg m2 = 3.3 kg Block 1 Block 2 ay 6.6 kg m-3:3 kg Case A Case B In Case A the mass of each block is 3.3 kg. In Case B the mass of block 1 (the block behind) is 6.6 kg, and the mass of block 2 is 3.3 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 6.5 N does act on block 2 in both cases and opposes the motion. For both Case A and Case B determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks. (a) Case AP = Case B P = (b) Case A3 - Case Ba

Answer 1

See the required free body diagram of both the bodies

As the blocks are movings towards right, but due to frcition on B, the acceleration will be towards left.

Acceleration of both the blocks has to be same, as they always remains in contact.

Case A:

given :

mass of A = mass of block 1 = m1 = 3.3 kg

mass of A = mass of block 1 = m2 = 3.3 kg

friction = f = 6.5 N.

a = acceleration of the blocks towards left = to be calculated.

N = Normal reaction = force with which the blocks pushes each other = to be calculated.

for A :

N = m1a ------ (i)

for B :

f - N = m2a ------ (ii)

from (i) and (ii)

f = (m1 + m2) a

=> a = f/(m1 + m2) = 6.5 / 6.6 = 0.98 m/s2.

also, N = m1a = 3.3 x 0.98 = 3.2 N.

Case B:

given :

mass of A = mass of block 1 = m1 = 6.6 kg

mass of A = mass of block 1 = m2 = 3.3 kg

friction = f = 6.5 N.

a = acceleration of the blocks towards left = to be calculated.

N = Normal reaction = force with which the blocks pushes each other = to be calculated.

for A :

N = m1a ------ (i)

for B :

f - N = m2a ------ (ii)

from (i) and (ii)

f = (m1 + m2) a

=> a = f/(m1 + m2) = 6.5 / 9.9 = 0.66 m/s2.

also, N = m1a = 6.6 x 0.66 = 4.4 N.

a) Case A, P = 3.2 N

Case B, P = 4.4 N

b) Case A, a = 0.98 m/s2.

Case B, a = 0.66 m/s2.