Question

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?

Answer 1

6x

Step-by-step explanation:

Answer 2

`x' = 10 x`

Step-by-step explanation:

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

so we will have

`(1)/(2)kx^2 = (1)/(2)mv^2`

so we will have

`v = \sqrt{(k)/(m)}(x)`

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

so we have

`5v = \sqrt{(k)/(4m)}(x')`

now from above two equations we have

`(5v)/(v) = (x')/(2x)`

so we have

`x' = 10 x`