An object of mass 9.9 kg is in equilibrium while connected to a light spring of

Question

An object of mass 9.9 kg is in equilibrium
while connected to a light spring of constant
158 N/m that is fastened to a wall. A second
object of mass 3.1 kg is slowly pushed up
against mass m1, compressing the spring by
the amount 0.26 m.
The system is then released, causing both
masses to start moving to the right on the
frictionless surface. When m1 is at the equi-
librium point, m2 loses contact with
m1 and moves to the right with speed v.
V= .91m/s.

How far apart are the objects when the spring
is fully stretched for the first time?
Answer in units of cm

Explanation / Answer

(1/2) m1 v2 = (1/2) k x12

0.5*9.9*(0.91)2 = 0.5*158*x12

>>>> x1 = 22.7788 cm

---------------------------------------------------------

it takes t seconds for the spring stretching:

t = T/4 = (2/)/4 = (2/(k/m))/4 = (/2) * ((m/k))

= (3.14/2) * ((9.9/158))

= 0.393 s

---------------------------------------------------------

displacement of the second mass is x2

x2 = v2 * t

     = 0.91 * 0.393

     = 35.763 cm

>>>> x2 = 35.763 cm

---------------------------------------------------------

x2 - x1 = 35.763 - 22.7788

           = 12.98 cm

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