Answer:
7.5 m/s² and 5 m/s²
Explanation:
Note: The force in the compressed spring is the same as the force needed to produce the acceleration in each box.
From Hook's law,
F = ke..................... Equation 1
Where F = force, k = spring constant, e = compression.
Given: k = 250 N/m, e = 6.0 cm = 0.06 m
Substitute into equation 1
F = 250×0.06
F = 15 N.
Using Newton's fundamental equation of kinematics
F = ma.................. Equation 2
Where m = mass of the box, a = acceleration of the box.
make a the subject of the equation
a = F/m................. Equation 3
For the first box,
Given: m = 2.0 kg, F = 15 N
Substitute into equation 3
a = 15/2.0
a = 7.5 m/s².
For the second box,
Give: m = 3.0 kg, F = 15 N
Substitute into equation 3
a = 15/3
a = 5 m/s²
The acceleration of the first box (with 2.0kg mass) is 7.5m/s² in one direction and the acceleration of the second box (with 3.0kg mass) is 5.0m/s² in the opposite direction.
The force (F) due to the compression (c) of the spring by both masses is given by Hooke's law as follows;
F = k x c -------------(i)
Where;
k = the spring's force constant.
From the question;
k = 250N/m
c = 6.0cm = 0.06m
Substitute these values into equation (i) as follows;
F = 250 x 0.06
F = 15N
This force causes both masses (boxes) to accelerate in opposite directions and the magnitude of this acceleration is given by Newton's second law of motion as follows;
F = m x a --------------(ii)
Where;
m = mass of each box
a = acceleration of each box
F = force causing the acceleration.
For the first box;
m = 2.0kg
Substitute this value into equation (ii) as follows;
F = 2.0 x a [F = 15N as calculated above]
15 = 2.0a
a = [tex]\frac{15}{2.0}[/tex]
a = 7.5m/s²
For the second box;
m = 3.0kg
Substitute this value into equation (ii) as follows;
F = 3.0 x a [F = 15N as calculated above]
15 = 3.0a
a = [tex]\frac{15}{3.0}[/tex]
a = 5.0m/s²
Therefore, the acceleration of the first box (with 2.0kg mass) is 7.5m/s² in one direction and the acceleration of the second box (with 3.0kg mass) is 5.0m/s² in the opposite direction.
