Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

A spring has K=24 N/m, we put in its end an object with 4 kg, if we pull the object 0.5 m then leave it, what would be the speed in 0.3 m from the balance point.

Sagot :

Answer:

Velocity- v=10m/s south Speed- v=10m/s  

Explanation:

the salar measurement of how fast an object is moving.

Velocity- v=10m/s south :

The vector rate of change of position.

Speed- v=10 m/s   :  the salar measurement of how fast an object is moving.

The velocity of the mass of 4 kg at 0.5 m is 1.732 m/sec. And the velocity of the mass of 4 kg at 0.3 m is 1.34 m/sec.

What is energy conversion?

The sum of kinetic energy and the potential energy of the body is constant. which is the energy of the remains conserved. This is known as energy conversion.

A spring has K=24 N/m.

We put in its end an object with 4 kg.

If we pull the object 0.5 m. Then the velocity will be given by the conservation of energy.

[tex]\begin{aligned} \rm K.E \ of \ mass &= \rm P.E. \ of \ spring\\\\ \dfrac{1}{2} m v^2 &= \dfrac{1}{2} kx^2\\\\\dfrac{1}{2}*4*v^2 &= \dfrac{1}{2} * 24 * 0.5\\\\v^2 &= 3\\\\v &= \sqrt{3} \\\\v &= 1.732 \ \ \rm m/sec \end{aligned}[/tex]

The speed at 0.3 m from the balance point. Then the velocity will be given by the conservation of energy.

[tex]\begin{aligned} \rm K.E \ of \ mass &= \rm P.E. \ of \ spring\\\\ \dfrac{1}{2} m v^2 &= \dfrac{1}{2} kx^2\\\\\dfrac{1}{2}*4*v^2 &= \dfrac{1}{2} * 24 * 0.3\\\\v^2 &= 1.8\\\\v &= \sqrt{1.8} \\\\v &= 1.34 \ \ \rm m/sec \end{aligned}[/tex]

More about the energy conservation link is given below.

https://brainly.com/question/2137260