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 acquires a maximum speed v and a maximum kinetic energy K. If instead the mass initially compresses the spring twice as far:______.
(a) its maximum speed will be 2v and its maximum kinetic energy will be 4K.
(b) its maximum speed will be 2v and its maximum kinetic energy will be √2K.
(c) its maximum speed will be 4v and its maximum kinetic energy will be 2K.
(d) its maximum speed will be 2v and its maximum kinetic energy will be 2K.
(e) its maximum speed will be v √2 and its maximum kinetic energy will be 2K.

Answer 1

Answer:

(a) its maximum speed will be 2v and its maximum kinetic energy will be 4K.

Explanation:

The velocity of the load if it were attached would be given by

$v=\omega\sqrt{A^2-x^2}$

ω is the angular velocity of the spring, A is the amplitude and x is the displacement of the load at any time.

When the spring is compressed (or extended), then the displacement, x, is maximum and equal to the amplitude, A. When released, at its equilibrium point, x is zero. It is at this point that the unattached load moves away.

Using x = 0 in the equation,

$v=\omega A$

This is the maximum velocity. ω depends on the mass of the load and the spring constant; hence it is constant.

It follows then that $v\propto A$.

When the compression is doubled, A is doubled. Because of the linear relationship between v and A, v is also doubled.

Kinetic energy is given by

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

When v is doubled, the kinetic energy becomes

$K_1 = \frac{1}{2} m(2v)^2 = 4(\frac{1}{2} mv^2) = 4K$

The kinetic energy is multiplied by a factor of 4.