Two blocks are sliding to the right across a horizontal surface, as the drawing

Question

Two blocks are sliding to the right across a horizontal surface, as the drawing shows.

In Case A the mass of each block is 3.4 kg. In Case B the mass of block 1 (the block behind) is 6.8 kg, and the mass of block 2 is 3.4 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 6.7 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.

Explanation / Answer

Let the blocks will accelerate towards the right with acceleration a.
Let F1 be the push force acting on the block 2 by block 1, then according to Newton's third law of motion, the same force will be acted on the block 1 by the block 2.

Case 1:
m1 = 3.4 kg ; m2 = 3.4 kg

The net forces acting on block 1,                       -F1 = m1a                       ...... (1)
The net forces acting on block 2,                      F1- f = m2a                       ...... (2)
Adding two equations:                          -f = ( m1 + m2 ) a
                           a = - 6.7 N / (6.8 kg)
                          a = -0.985 m / s2 (blocks are getting slower)
The push force                            F1 = ( -3.4 kg )( -0.98 m/s2 )
                                = 3.332 N _______________________________________________________ _______________________________________________________ Case 2:
m1 = 6.8 kg ; m2 = 3.4 kg The net forces acting on block 1,                       -F1 = m1a                       ...... (1)
The net forces acting on block 2,                      F1- f = m2a                       ...... (2) Adding two equations:                          -f = ( m1 + m2 ) a                           a = - 6.7 N / (10.2  kg)
                          a = -0.6568   m / s2 (blocks are getting slower)
The push force                            F1 = ( -6.8 kg )( -0.6568  m/s2 )
                                = 4.46 N
Adding two equations:                          -f = ( m1 + m2 ) a                           a = - 6.7 N / (10.2  kg)
                          a = -0.6568   m / s2 (blocks are getting slower)
The push force                            F1 = ( -6.8 kg )( -0.6568  m/s2 )
                                = 4.46 N
                          a = - 6.7 N / (10.2  kg)
                          a = -0.6568   m / s2 (blocks are getting slower)
The push force                            F1 = ( -6.8 kg )( -0.6568  m/s2 )
                                = 4.46 N                            F1 = ( -6.8 kg )( -0.6568  m/s2 )
                                = 4.46 N

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