Two blocks are in contact on a frictionless table. A horizontal force F is applied to M2, as shown. If M1 = 1.06 kg, M2 = 3.80 kg, and F = 4.85 N, find the size of the contact force between the two blocks.
If instead an equal but oppositely directed force is applied to M1 rather than M2, find the size of the contact force between the two blocks.
F=>M2=>F1=>M1
On the frictionless table, both will have an acceleration a.
Draw the free-body-diagrams (FBD) and conclude that
M2.a = (F-F1)
M1.a = F1
using Newton's second law.
Solve for F1 and a.
Part (b) can be found in a similar way.
To find the size of the contact force between the two blocks, we need to consider the forces acting on each block individually.
1. When force F is applied to M2:
Since the table is frictionless, the only forces acting on M2 are the applied force F and the contact force between the two blocks.
The net force on M2, according to Newton's second law, is given by:
F_net = M2 * a
where M2 is the mass of M2, and a is the acceleration of the system.
We can find the acceleration of the system using the applied force and the mass of both blocks:
F_net = (M1 + M2) * a
Rearranging the equation, we have:
a = F_net / (M1 + M2)
Now, we can find the contact force between the two blocks using the acceleration:
Contact force = M1 * a
Substituting the given values:
M1 = 1.06 kg
M2 = 3.80 kg
F = 4.85 N
F_net = F - frictional force (since the table is frictionless, the frictional force is zero)
Plugging in the values, we have:
F_net = 4.85 N - 0 (frictional force) = 4.85 N
a = F_net / (M1 + M2) = 4.85 N / (1.06 kg + 3.80 kg) = 4.85 N / 4.86 kg ≈ 0.998 m/s²
Contact force = M1 * a = 1.06 kg * 0.998 m/s² ≈ 1.058 N
Therefore, the size of the contact force between the two blocks when force F is applied to M2 is approximately 1.058 N.
2. When an equal but oppositely directed force is applied to M1:
In this case, the analysis is similar to the previous case, but the roles of M1 and M2 are reversed.
The net force on M1 is given by:
F_net = M1 * a
We can find the acceleration using the applied force and the mass of both blocks:
a = F_net / (M1 + M2)
The contact force between the two blocks is then:
Contact force = M2 * a
Substituting the given values:
M1 = 1.06 kg
M2 = 3.80 kg
F = 4.85 N
F_net = F - frictional force (since the table is frictionless, the frictional force is zero)
Plugging in the values, we have:
F_net = -4.85 N - 0 (frictional force) = -4.85 N
a = F_net / (M1 + M2) = -4.85 N / (1.06 kg + 3.80 kg) = -4.85 N / 4.86 kg ≈ -0.999 m/s²
Contact force = M2 * a = 3.80 kg * -0.999 m/s² ≈ -3.797 N
Therefore, the size of the contact force between the two blocks when an equal but oppositely directed force is applied to M1 is approximately -3.797 N (negative because the direction is opposite).
Note: The negative sign indicates that the contact force is in the opposite direction to the applied force.
To find the size of the contact force between the two blocks, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's break down the problem into two scenarios:
Scenario 1: When the force is applied to M2.
In this case, the force F is applied to M2. Since there is no friction involved, the force F will accelerate both M2 and M1. The contact force between the two blocks will be the same as the force applied to M1.
To find the acceleration of the system, we can use the following equation:
F = (M1 + M2) * a
Rearranging the equation, we get:
a = F / (M1 + M2)
Substituting the given values:
a = 4.85 N / (1.06 kg + 3.80 kg)
Calculating the acceleration:
a = 4.85 N / 4.86 kg
a ≈ 1.00 m/s^2
Since both blocks are accelerating at the same rate, the contact force between the two blocks is equal to the force applied to M1:
Contact force = Force applied = 4.85 N
Scenario 2: When the force is applied to M1.
In this case, the equal but oppositely directed force is applied to M1. Again, there is no friction involved. Since the force acts in the opposite direction, the acceleration of the system will be different compared to Scenario 1.
Using the same equation as before to find the acceleration:
F = (M1 + M2) * a
Rearranging the equation, we get:
a = F / (M1 + M2)
Substituting the given values:
a = 4.85 N / (1.06 kg + 3.80 kg)
Calculating the acceleration:
a = 4.85 N / 4.86 kg
a ≈ 1.00 m/s^2
Once again, since both blocks are accelerating at the same rate, the contact force between the two blocks is equal to the force applied to M2:
Contact force = Force applied = 4.85 N
Therefore, regardless of whether the force is applied to M2 or M1, the size of the contact force between the two blocks is 4.85 N.