A mass is pressed against (but is not attached to) an ideal horizontal spring on a frictionless horizontal surface. After being released from rest, the mass aquires a max speed v and amax kinetic energy K. If instead the mass intially compresses the spring twice as far. A- Its max speed will be 2v and it's max kenitic energy will be Sqroot(2) K. B- Its max speed will be 4v and it's max kenitic energy will b 2 K. C- Its max speed will be v Sqroot(2) and it's max kenitic energy will b 2 K. D- Its max speed will be 2v and it's max kenitic energy will b 4 K. E- Its max speed will be 2v and it's max kenitic energy will b 2 K.

D. Its max speed will be 2v and it's max kinetic energy will be 4K

Explanation:

Let is the mass of object, is the spring constant of spring and initially it compresses the spring by meters.

As there is no friction so conservation of energy will be followed. So,

Potential energy stored in the spring = Kinetic energy acquired by the object

                  (Equation 1)

where K is the maximum kinetic energy of the object.

Again we can write as

So,              (Equation 2)

According to the question, the mass is compressing the sprint twice than before, so new compression will be and we can write

  (from equation 1)

So the new kinetic energy of the mass will be 4K.

Again,

So,

From equation 1 we can put the value of and thus we write

Thus the new speed of the mass will be 2v.

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.