Use the approximation that for each time step. A spring with a relaxed length of
ID: 2306365 • 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 72 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 30.9 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.)
= <0,.08227,0> X
=<0,.005923,0> X
I've worked these out, but the computer says they're wrong. What's wrong?
r
= <0,.08227,0> X
p
=<0,.005923,0> X
Explanation / Answer
the length of the spring after stretched = 30.9 cm so stretched lenght = 30.9 - 25 = 5.9 cm = x
when the block is stretched up wards two froces are acting on the block downwards ,
one is force due to spring Fs , and 2nd is gravity Fg ; so the net force Fnet = Fs+ Fg
Fs = k x = 14 N / m * 0.059 m= 0.826 N
and Fg = m * g = 0.072 kg * 9.8 m/sec^^2 = 0.705
Fnet = Fg+ Fs = 0.826 + 0.705 = 1.531 N this force is acting on the block verticall downwards
when the block was just released the acceleration experienced by the block just after release a = Fnet / m
a = 1.531 / 0.072 kg = 21.263 m / sec^^2
let S be distance travelled by the block in time t = 0.2 sec ;
S = at^2 / 2 = 21.263 ( m / sec^2 ) (0.2 ^2) / 2
= 0.425 m or 42.0cm
momentum of block p = mv; v = a t = 21.263*0.2s = 4.252 m/sec
p = 0.072 kg *4.252 m/sec = 0.306 kg m/sec along y axis vertically downword
in vector form p = -0.306 j
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