A 3.39 kg block is placed on top of a 11.8 kg

block. A horizontal force of F = 61.3 N is
applied to the 11.8 kg block, and the 3.39 kg
block is tied to the wall. The coefficient of
kinetic friction between all moving surfaces
is 0.112. There is friction both between the
massesandbetweenthe11.8kgblockandthe
ground.
The accelerationof gravityis9.8m/s2.

To find the acceleration of the system, we need to first calculate the net force acting on the system.

1. Calculate the force of gravity acting on each block:
- The force of gravity (weight) acting on the 3.39 kg block (m1) is given by: F1 = m1 * g
- The force of gravity acting on the 11.8 kg block (m2) is given by: F2 = m2 * g
where g is the acceleration due to gravity (9.8 m/s^2).

Substituting the given values:
F1 = 3.39 kg * 9.8 m/s^2 = 33.222 N
F2 = 11.8 kg * 9.8 m/s^2 = 115.64 N

2. Calculate the frictional force acting between the 11.8 kg block and the ground:
The frictional force (Ff1) between the 11.8 kg block and the ground is given by: Ff1 = μ * N
where μ is the coefficient of kinetic friction and N is the normal force.

The normal force (N) is equal to the force of gravity acting on the 11.8 kg block: N = F2 = 115.64 N

Substituting the given coefficient of kinetic friction:
Ff1 = 0.112 * 115.64 N = 12.96068 N

3. Calculate the net force on the system:
The net force (Fnet) on the system is the horizontal force applied minus the frictional force: Fnet = F - Ff1

Substituting the given applied force:
Fnet = 61.3 N - 12.96068 N = 48.33932 N

4. Calculate the acceleration of the system:
The acceleration (a) of the system is equal to the net force divided by the total mass of the system: a = Fnet / (m1 + m2)

Substituting the given masses:
a = 48.33932 N / (3.39 kg + 11.8 kg) = 48.33932 N / 15.19 kg = 3.182 m/s^2

Therefore, the acceleration of the system is 3.182 m/s^2.