Use the approximation that for each time step. A spring with a relaxed length of
ID: 1838700 • Letter: U
Question
Use the approximation that
for each time step.
A spring with a relaxed length of 25 cm and a stiffness of 14 N/m stands vertically on a table. A block of mass 73 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 30.0 cm, hold the block at rest for a moment, and then release it. Using a time step of 0.1 s, predict the position and momentum of the block at a time 0.2 s after you release the block. (Assume the +y direction is upward. Express your answers in vector form.)
Explanation / Answer
The force on such type of system
F = kx + mg
ma = kx+mg
initial value of x = 30-25 = 5cm
a = k/m *x + g = 14/(73*10^(-3)) * 5*10^(-2) + 9.81 = 19.34 m/s2
Ist step:
s = ut + 1/2 at2
=0 + .5*19.34*(.1)2 = 0.0967 m = 9.67 cm
thus after 0.1 sec it will at 30-9.67 = 20.33 below
now at this time it is below rest value lengh of spring
now at this time force equation
ma = -kx + mg
a = -14/73 (25-20.33)*10^(-1) + 9.81 = 0.854 m/s2
speed v = u + at = 0+19.34*0.1 = 1.934 m/s
2nd step:
s = ut + 1/2 at2
u is not zero because object in motion
s = 1.934 * 0.1 + 1/2 * 0.854 * 0.12
= 0.198 m
now it will below from the equlibirium position
20.33-19.8 = 0.53 cm
vector R = 0.53 j
vector p = m * vector v where v = u+ at = 1.934 + 0.854*0.1 = 2.01 m/s
vector p = - 73 * 2.01 j = -146.73 j
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