Use the approximation that v→avg=p→f/m for each time step. A spring with a relaxed length of 25 cm and a stiffness of 16 N/m stands vertically on a table. A block of mass 89 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 30.3 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.) r→= < , , > m p→= < , , > kg·m/s
We have to solve the differential equation:
Now . Therefore the equation to solve is
. Initially at time t=0,x=30.3-25=5.3 cm=5.3*10-2 m.
At any time t, if is very small we can express the position at time in terms of the position and velocity at time t to a good approximation as . Also .
Now and x(0)=5.3*10-2 m.
Therefore x(0.1)=x(0)+0.1*v(0)=5.3*10-2 m since v(0)=0 m/s.
v(0.1)=v(0)-0.1*180*x(0)=0-18*x(0)=-0.18*5.3*10-2=-0.95 m/s.
At t=0.2 s,
To determine the momentum of the block we only have to use the relation p=mv.
Use the approximation that v→avg=p→f/m for each time step. A spring with a relaxed length of...
Use the approximation that v→avg=p→f/m for each time step. A spring with a relaxed length of 25 cm and a stiffness of 16 N/m stands vertically on a table. A block of mass 89 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 30.3 cm, hold the block at rest for a moment, and then release it. Using a time step of 0.1 s, predict the position...
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