Q6-39E

Question

A $6 .0 kg$ box moving at $3 .0 m/s$ on a horizontal, frictionless surface runs into a light spring of force constant $75 .0 N/cm$ . Use the work-energy theorem to find the maximum compression of the spring.

Step-by-Step Solution

Verified

Answer

The maximum compression in the spring is $8.5 cm$ .

1Step 1: Identification of given data

The given data can be listed below,

The mass of the box is, $m = 6.0 kg$ .
The velocity of the box is, $v = 3.0 m/s$ .
The force constant of spring is, $k = 75 N/cm$ .

2Step 2: Concept/Significance of work energy theorem

The work-energy equation, which shows a direct correlation between a particle's velocity and the net work performed on it.

3Step 3: Determination of the maximum compression of the spring

The total work done on the box is given by,

$\begin{matrix} W_{t o t} = K_{f} - K_{i} \\ = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} \end{matrix}$

…(i)

Here, m is the mass of the box and $v_{f}$ is the final velocity of the box whose value is zero as it comes to rest in last, $v_{1}$ is the initial velocity of the box.

Substitute all the values in the above,

$\begin{matrix} W_{t o t} = \frac{1}{2} m {(0)}^{2} - \frac{1}{2} m {(3.0 m/s)}^{2} \\ = - \frac{1}{2} (6.0 kg) {(3.0 m/s)}^{2} \\ = - 27 J \end{matrix}$

According to the conservation law, the work done due to spring force and the total work done are equal. The work done due to spring force is given by,

$W = - \frac{1}{2} k x^{2}$ …(ii)

Here, k is the force constant of the spring, $x$ is the compressed length of the spring.

Substitute all the values in the above,

$\begin{matrix} - 27 J = - \frac{1}{2} (75 N/cm) x^{2} \\ x = \sqrt{\frac{2 \times 27 J}{75 N/cm}} \\ = 8.5 cm \end{matrix}$