A 1.30kg kg box and a 3.80kg kg box on a perfectly smooth horizontal floor have a spring of force constant 300N/m N/m compressed between them. The initial compression of the spring is 5.00cm cm .
Part A
Find the acceleration of 1.30kg kg box the instant after they are released.
Part B
Find the acceleration of 3.80kg kg box the instant after they are released.
To find the acceleration of each box, we can use Newton's second law of motion, which states:
F = ma
where F is the net force acting on the object, m is the mass of the object, and a is the acceleration.
Part A:
Let's start by finding the net force acting on the 1.30 kg box.
The net force is equal to the force exerted by the spring. The force exerted by a spring can be calculated using Hooke's Law, which states:
F = -kx
where F is the force exerted, k is the force constant (spring constant), and x is the displacement from the equilibrium position.
In this case, the spring is compressed by 5.00 cm, which is equivalent to 0.05 m. The force constant is given as 300 N/m. So, the force exerted by the spring is:
F = -kx
F = -(300 N/m)(0.05 m)
F = -15 N
The net force acting on the box is equal to the force exerted by the spring. Since the direction of the force is opposite to the direction of positive x-axis, we consider it as negative.
Now, we can find the acceleration of the 1.30 kg box using Newton's second law:
F = ma
-15 N = (1.30 kg) * a
Solving for a:
a = -15 N / 1.30 kg
a ≈ -11.54 m/s^2
So, the acceleration of the 1.30 kg box, instant after it is released, is approximately -11.54 m/s^2.
Part B:
To find the acceleration of the 3.80 kg box, we need to consider the net force acting on it as well.
The force exerted by the spring is the same as in Part A, which is -15 N.
Using Newton's second law:
F = ma
-15 N = (3.80 kg) * a
Solving for a:
a = -15 N / 3.80 kg
a ≈ -3.95 m/s^2
Therefore, the acceleration of the 3.80 kg box, instant after it is released, is approximately -3.95 m/s^2.