Question

A spring with a force constant of 100.0 N/m is used to send a box sliding across a horizontal surface. The box, with amass of 55.0 g, is pushed against the horizontal spring, compressing it 11.0 cm from equilibrium. The system is thenreleased.a) What is the speed of the box at the instant the spring is passing the point of being compressed 7.0 cm? (It has nowmoved 4 cm away from the point of maximum compression.

Answer 1

Given:

The force constant of a spring is k = 100 N/m

The mass of the box is 55 g = 0.055 kg

The maximum compression is

$\begin{gathered} x_{max}=\text{ 11 cm} \\ =0.11\text{ m} \end{gathered}$

Required:

Speed of the box when it is compressed to x = 7 cm = 0.07 m

Explanation:

The speed of the box can be calculated as

$\begin{gathered} v=\sqrt{\frac{k}{m}}\times\sqrt{(x_{max}^2-x^2)} \\ =\sqrt{\frac{100}{0.055}}\times\sqrt{((0.11)^2-(0.07)^2)} \\ =3.618\text{ m/s} \end{gathered}$

Final Answer: The speed of the box is 3.618 m/s.