Question

Block 1 (m1 = 7 kg) and block 2 (m2 = 9 kg) are adjacent to each other on the surface of a table. Block 2 is to the LEFT of block 1. A hand pushes horizontally on block 2 and causes both blocks to move to the right with an acceleration of magnitude 2 m/s². The coefficient of kinetic friction between the block and the surface equals 0.4.
On a sheet of paper, draw the free body diagrams for block 1 and block 2 using the two-subscript notation from class. After completing the free body diagrams, enter below each force and its x & y-components. Remember that the x-component is the "i" component and the y-component is the "j" component.

NET force on Block 1
F_net1 = 14  i + 0  j N

NET force on Block 2
F_net2 = 18  i + 0  j N

FORCES on BLOCK 1
Weight force on block 1 by Earth  
W1E =  i +  j N

Remember that W1E points DOWN (negative y-value) with a magnitude of m1 * g.

Normal force on block 1 by Surface  
N1S =  i +  j N

Since the block does not move vertically, the upward N1S force must cancel the downward W1E force.

Frictional force on block 1 by Surface  
f1S =  i +  j N

The kinetic friction on block 1 by the surface (f1S) equals the kinetic coefficient of friction times the normal force between block 1 and the surface (N1S). In this particular case with another external force acting on the block (N12), the kinetic friction opposes the direction of motion.

Normal force on block 1 by block 2  
N12 =  i +  j N

In the horizontal direction, the sum of the N12 and f1S forces must equal the net force on block 1.

FORCES on BLOCK 2
Weight force on block 2 by Earth  
W2E =  i +  j N

Normal force on block 2 by Surface  
N2S =  i +  j N

Frictional force on block 2 by Surface
f2S =  i +  j N

Normal force on block 2 by block 1  
N21 =  i +  j N

Normal force on block 2 by Hand
N2H =  i +  j N

Can you help me with this?

Answer 1

a)

N

b)

N

c)

N

d)

e)

N

f)

N

g)

h)

i)

j)

N

k)