Question

A 2.0-kg box and a 3.0-kg box on a perfectly smooth horizontal floor have a spring of force constant 250 N/m compressed between them. If the initial compression of the spring is 6.0 cm, find the acceleration of each box the instant after they are released

Answer 1

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

Answer 2

Answer:

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.

Explanation:

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 = $\frac{15}{2.0}$

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 = $\frac{15}{3.0}$

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.