Three blocks are in contact with each other
on a frictionless horizontal surface. A 696 N
horizontal force is applied to the block with
mass of 4.7 kg as shown in the figure below.
The acceleration of gravity is 9.8 m/s
2
.
4.7 kg 5.5 kg 7.1 kg
) What is the resultant force on the block
with mass 5.5 kg?
Answer in units of N
To find the resultant force on the block with a mass of 5.5 kg, we need to first determine the individual forces acting on each block and then apply Newton's second law of motion.
1. Calculate the force applied to the 4.7 kg block:
The force applied to the 4.7 kg block is given as 696 N.
2. Find the acceleration of the system:
To find the acceleration, we will use Newton's second law of motion:
Force = mass * acceleration
The total mass of the system is the sum of the masses of all three blocks: 4.7 kg + 5.5 kg + 7.1 kg.
The total force acting on the system is 696 N.
So, using Newton's second law, we can calculate the acceleration:
696 N = (4.7 kg + 5.5 kg + 7.1 kg) * acceleration
Simplifying the equation, we get:
696 N = 17.3 kg * acceleration
Solving for acceleration, we get:
acceleration = 696 N / 17.3 kg
3. Calculate the force on the block with a mass of 5.5 kg:
Now that we have the acceleration, we can find the force on the 5.5 kg block using Newton's second law:
Force = mass * acceleration
Force = 5.5 kg * acceleration
Substituting the value of acceleration we calculated earlier, we get:
Force = 5.5 kg * (696 N / 17.3 kg)
Calculating the value, we get:
Force = 221.82 N
Therefore, the resultant force on the block with a mass of 5.5 kg is 221.82 N.