wo blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block A has mass 14.5 kg , and block B has mass m. A constant horizontal force F = 60.0 N is applied to block A (Figure 1). In the first 5.00 s after the force is applied, block A moves 18.0 m to the right.
To find the mass of block B, we can use Newton's second law of motion.
The formula for Newton's second law is:
F = ma
where F is the net force acting on an object, m is the mass of the object, and a is the acceleration of the object.
In this case, block A is experiencing a constant horizontal force of 60.0 N. We know that block A has a mass of 14.5 kg and moves 18.0 m to the right in the first 5.00 seconds.
First, let's find the acceleration of block A using the formula:
a = (change in velocity) / (time)
The change in velocity is the final velocity minus the initial velocity. Since block A is at rest initially, the initial velocity is 0 m/s. The final velocity can be calculated by dividing the displacement by the time:
v = d / t
where v is the velocity, d is the displacement, and t is the time.
Plugging in the values, we have:
v = 18.0 m / 5.00 s = 3.60 m/s
Now, the acceleration of block A is:
a = (3.60 m/s - 0 m/s) / 5.00 s = 0.72 m/s^2
Using Newton's second law, we have:
F = ma
Substituting the values:
60.0 N = 14.5 kg * 0.72 m/s^2
Now, we can solve for the mass of block B:
m = (60.0 N) / (0.72 m/s^2) = 83.33 kg
Therefore, the mass of block B is approximately 83.33 kg.