Question

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.

Answer 1

Answer:

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

Explanation:

Let $m$ is the mass of object, $k$ is the spring constant of spring and initially it compresses the spring by $x$ 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

$\frac{1}{2} kx^2 = K$                   (Equation 1)

where K is the maximum kinetic energy of the object.

Again we can write as

$\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

So, $v=\sqrt{\frac{kx^2}{m} }$              (Equation 2)

According to the question, the mass is compressing the sprint twice than before, so new compression will be $2x$ and we can write

$\frac{1}{2} k(2x)^2 = 4\times \frac{1}{2} kx^2 = 4K$   (from equation 1)

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

Again,

$4\times \frac{1}{2} kx^2 = \frac{1}{2} mv_2^2$

So, $v_2= \sqrt{\frac{4\times kx^2}{m} }$

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

$v_2=\sqrt{4v} = 2v$

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

Answer 2

Answer:D

Explanation: