Question

6E. 01.222 J8. An object of mass 0.75 kg is brought in contactwith a spring of Hook's constant 220 N/m that iscompressed by 0.175 m. If the spring is let go free to expand,calculate the speed by which the object willleave the spring at its relaxed position. (1 point)A. O2.997 m/sB. O2457msC. 00.368 m/sD. O 4.427 m/sE. O5.221 m/s9. If all the forces acting onobiect arean

Answer 1

Given:

The spring constant of the spring is k = 220 N/m

The compression is x = 0.175 m

The mass of the object is m = 0.75 kg

To find the speed at which an object will leave the spring.

Step-by-step explanation:

According to the conservation energy, elastic potential energy is equal to kinetic energy.

The speed can be calculated by the formula

`\begin{gathered} (1)/(2)kx^2=(1)/(2)mv^2 \\ v=\sqrt{(kx^2)/(m)} \end{gathered}`

On substituting the values, the speed will be

`\begin{gathered} v=\sqrt{(220*(0.175)^2)/(0.75)} \\ =2.997\text{ m/s} \end{gathered}`

Thus, the speed is 2.997 m/s