In Case A the mass of each block is 4.8 kg. In Case B the mass of block 1 (the block behind) is 9.6 kg, and the mass of block 2 is 4.8 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 9.4 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) In Case A, since the mass of each block is the same (4.8 kg), the magnitude of the forces with which the blocks push against each other is equal, right around the "we're in this together" level of around 4.8 kg.
In Case B, things are a little more complicated. The mass of block 1 is double that of block 2, so block 1 definitely has more "pushing power." However, block 2 has the advantage of a cool kinetic frictional force of 9.4 N opposing its motion. So, even though block 1 has more mass, block 2 isn't going to back down so easily. The magnitude of the forces with which the blocks push against each other in Case B would be somewhere between "double teaming" and "you shall not pass" levels.
(b) To determine the magnitude of the acceleration of the blocks in both cases, we can use Newton's second law (F = ma). Let's break it down pointe-by-pointe.
In Case A, since the forces with which the blocks push against each other are equal, there is no net force acting on the system. Therefore, the total mass of the blocks (4.8 kg + 4.8 kg) times the acceleration would be equal to zero. So, the magnitude of the acceleration would be zero. In simpler terms, the blocks are going to enjoy some quality time standing still together.
In Case B, it gets a tad bit more interesting. With the kinetic frictional force of 9.4 N opposing block 2's motion, that force will slightly reduce the acceleration. However, it won't stop the determined duo of blocks completely. The magnitude of the acceleration in Case B would be a nonzero value, greater than zero (but not too great because of that sneaky frictional force), symbolizing a "no retreat, no surrender" attitude of the blocks.
So, to summarize:
(a) In Case A, the magnitude of the forces with which the blocks push against each other is 4.8 kg. In Case B, it's somewhere between "double teaming" and "you shall not pass" levels.
(b) In Case A, the magnitude of the acceleration is zero. In Case B, it's a nonzero value, symbolizing a "no retreat, no surrender" attitude.
To determine the magnitude of the forces with which the blocks push against each other and the magnitude of the acceleration of the blocks, we can use Newton's second law of motion and apply it to each block separately.
(a) Magnitude of the forces with which the blocks push against each other:
In both cases, the force with which block 1 pushes on block 2 is equal in magnitude and opposite in direction to the force with which block 2 pushes on block 1. Let's denote this force as F_push.
In Case A, since both blocks have the same mass of 4.8 kg and no frictional force acts on block 1, the force of kinetic friction on block 2 is the same as the force with which block 1 pushes on block 2. Therefore, the magnitude of the force with which the blocks push against each other is 9.4 N.
In Case B, the mass of block 1 is 9.6 kg and the mass of block 2 is 4.8 kg. Again, the force of kinetic friction opposing the motion of block 2 is the same as the force with which block 1 pushes on block 2. However, since block 1 has twice the mass as in Case A, the magnitude of the force with which the blocks push against each other will also be doubled. Therefore, the magnitude of the force with which the blocks push against each other in Case B is 18.8 N.
(b) Magnitude of the acceleration of the blocks:
To find the magnitude of the acceleration, we can use Newton's second law:
F_net = ma
In both cases, the only external force acting on the system is the force of kinetic friction acting on block 2. Therefore, the net force on the system is equal to this frictional force, and we can write:
F_net = 9.4 N (Same for both cases)
For Case A:
F_net = ma
9.4 N = 4.8 kg * a
Solving for the acceleration:
a = 9.4 N / 4.8 kg
a ≈ 1.958 m/s²
For Case B:
F_net = ma
9.4 N = (9.6 kg + 4.8 kg) * a
Solving for the acceleration:
a = 9.4 N / 14.4 kg
a ≈ 0.653 m/s²
So, the magnitude of the acceleration in Case A is approximately 1.958 m/s², and the magnitude of the acceleration in Case B is approximately 0.653 m/s².
To determine the magnitude of the forces with which the blocks push against each other in both Case A and Case B, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
In Case A, since both blocks have the same mass of 4.8 kg and no frictional force acts on block 1, the only force acting on block 2 is the kinetic frictional force of 9.4 N. Thus, the net force on block 2 is also 9.4 N. Since the blocks are in contact and push against each other, the force with which the blocks push against each other is equal in magnitude but opposite in direction to the net force on block 2. Therefore, the force with which the blocks push against each other in Case A is 9.4 N.
In Case B, block 1 has a mass of 9.6 kg and block 2 has a mass of 4.8 kg. Both blocks are subjected to a kinetic frictional force of 9.4 N. Since the kinetic frictional force acts in the opposite direction of the applied force, we can subtract it from the net force to find the force with which the blocks push against each other.
For block 1, the net force is given by:
net force = (mass of block 1) * acceleration of blocks
For block 2, the net force is given by:
net force = (mass of block 2) * acceleration of blocks + kinetic frictional force
Since the blocks are in contact and push against each other, the force with which they push against each other is equal in magnitude but opposite in direction to the net force on block 1. Therefore, the force with which the blocks push against each other in Case B is equal to the net force on block 1.
To find the acceleration of the blocks in both cases, we can rearrange the equations above. From the equation for block 1:
acceleration of blocks = net force / (mass of block 1)
From the equation for block 2:
acceleration of blocks = (net force - kinetic frictional force) / (mass of block 2)
Now we can substitute the given values into the equations to find the force with which the blocks push against each other and the acceleration of the blocks in both cases.
Assume that the blocks accelerate to the right with acceleration ‘a’
F1 is the force acting on the block 2 by block 1, and the same force (according to the 3 Newton's Law) acts on the block 1 by the block 2.
(A) m1=m2=4.8 kg
m1•a=-F1
m2•a=F1-F(fr).
Adding two equations
a• (m1+m2)= F(fr)
a=F(fr)/(m1+m2) = 9.4/9.6 =0.98 m/s²
Net force acting on the block 1
F1= - m1•a= - 4.8•0.98= - 4.7 N
The net forces acting on block 2
F1- F(fr) = m2•a = 4.8•0.98=4.7 N
(B)
m1•a=-F1
m2•a=F1-F(fr)
Adding two equations
a• (m1+m2)= F(fr)
a=F(fr)/(m1+m2) = 9.4/9.6 =0.65 m/s²
Net force acting on the block 1
F1=-m1•a=4.8•0.6.5=3.12 N
The net forces acting on block 2
F1- F(fr) = m2•a = 9.6•0. 65=6.24 N