Relaxed : length Push down release from rest A spring has a relaxed length of 33
ID: 3278883 • Letter: R
Question
Relaxed : length Push down release from rest A spring has a relaxed length of 33 cm (0.33 m) and its spring stiffness is 12 N/m. You glue a 70 gram block (0.070 kg) to the top of the spring, and push the block down, compressing the spring so its total length is 16 cm. You make sure the block is at rest, then at time t 0 you quickly move your hand away. The block begins to move upward, because the upward force on the block by the spring is greater than the downward force on the block by the Earth. Calculate approximately y vs. time for the block during a 0.24-second interval after you release the block, by applying the Momentum Principle in three steps each of 0.08-second duration.Explanation / Answer
Here is a small self explainatory MATLAB Code which simulated the whole situation using Momentum principle:---
r in the code is "y" of the question
%% *********************CODE ---- all quantities in S.I. Units**********************
k=12;
dx0 = 0.17;
m=0.07;
g=9.8;
dt=0.08;
p0=0;
r0= 0.16;
v0=0;
f0=0;
for i=1:3
f(i)=k*(dx0)-m*g;
p(i)= p0+f(i)*dt;
v(i)=p(i)/m;
r(i)= r0+(v(i)+0)*dt;
p0=p(i);
r0=r(i);
dx0= 0.33-(r(i));
f0=f(i);
end
%% **************************************************
Calculations:---
for i =1 ==> t = 1 x dt = 0.08 s
f(i)=k*(dx0)-m*g
f =
1.3540N ;
p(i)= p0+f(i)*dt
p =
0.1083 kg-m/s
v(i)=p(i)/m
v =
1.5474 m/s
r(i)= r0+(v(i)+0)*dt
r =
0.2838 m
*******************************
for i =2 ==> t = 2 x 0.08 = 0.16 s
f(i)=k*(0.33-0.2838)-m*g
f =
-0.1315N ;
p(i)= p(t= 0.08)+f*dt
p =
0.0978 kg-m/s
v(i)=p/m
v =
1.397 m/s
r(i)= 0.2838+(1.397)*dt
r =
0.3956 m
*******************************
Similarly for i =3 ==> t = 3 x 0.08 = 0.24 s
f= -1.4728 N ;
f (spring) = k(0.33-0.3956) = -0.79 N
f(earth) = mg = -0.686 N
v = -0.286 m/s
p= -0.02 kg-m/s
r = 0.3727 m.