Answer :
a. Free diagrams of each block b. The force of tension in the string is approximately 0.7213 N. c. The acceleration of Block B is 0 m/s²
a. Free Body Diagrams
Block A:
- Force of tension (T) in the string, directed towards the right.
- Force due to friction (f) acting towards the left, opposing the motion.
- Force of gravity (mg) acting downwards.
- Normal force (N) exerted by the surface, directed upwards.
Block B:
- Force of tension (T) in the string, directed towards the right.
- Force of friction (f) acting towards the left, opposing the motion.
- Force of gravity (mg) acting downwards.
- Normal force (N) exerted by the surface, directed upwards.
b. To determine the force of tension in the string, we need to analyze the forces acting on Block A.
Since Block A is in equilibrium and not accelerating vertically, the vertical forces cancel out, and we only need to consider the horizontal forces.
The force of friction can be calculated using the equation:
f = μN
where μ is the coefficient of friction and N is the normal force.
The normal force (N) on Block A is equal to the weight of Block A, which is given by:
N = mg
Substituting the values:
N = (2.3 kg)(9.8 m/s^2)
N ≈ 22.54 N
Now, we can calculate the force of friction:
f = μN
f = (0.0320)(22.54 N)
f ≈ 0.7213 N
Since the force of tension in the string is equal in magnitude but opposite in direction to the force of friction, the force of tension is:
T = f ≈ 0.7213 N
Therefore, the force of tension in the string is approximately 0.7213 N.
c. To determine the acceleration of Block B, we need to analyze the forces acting on it
Since Block B is moving horizontally, we need to consider the net force acting on it.
The net force on Block B in the horizontal direction is given by:
Net Force on B = T - f
To calculate the force of friction, we use the equation:
f = μN
where μ is the coefficient of friction and N is the normal force.
The normal force (N) on Block B is equal to the weight of Block B, which is given by:
N = mg
Substituting the values:
N = (51.7 kg)(9.8 m/s^2)
N ≈ 506.66 N
Now, we can calculate the force of friction:
f = μN
f = (0.0320)(506.66 N)
f ≈ 16.213 N
The net force on Block B is:
Net Force on B = T - f
To determine the acceleration of Block B, we can use Newton's second law:
Net Force on B = mB * a
Where mB is the mass of Block B and a is its acceleration.
Substituting the values and solving for a:
T - f = mB * a
T - f = (51.7 kg) * a
Since the force of tension is equal to the force of friction (T = f), we have:
0 = (51.7 kg) * a
This implies that the net force on Block B is zero, resulting in no acceleration. Therefore, Block B does not accelerate and remains stationary.
Hence, the acceleration of Block B is 0 m/s².
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