contestada

A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed of 5v. What distance was the spring compressed in the second case?

Respuesta :

Answer:

[tex]x' = 10 x[/tex]

Explanation:

By energy conservation we know that spring energy is converted into kinetic energy of the block

so we will have

[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]

so we will have

[tex]v = \sqrt{\frac{k}{m}}(x)[/tex]

now we will have same thing for another mass 4m which moves out with speed 5v

so we have

[tex]5v = \sqrt{\frac{k}{4m}}(x')[/tex]

now from above two equations we have

[tex]\frac{5v}{v} = \frac{x'}{2x}[/tex]

so we have

[tex]x' = 10 x[/tex]

Otras preguntas