Use the approximation that vavg pfm for each time step A sp

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 13 N/m stands vertically on a table. A block of mass 78 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 29.7 cm, hold 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 middot m/s

Solution

equilibrium positon for spring,

ky - m g = 0

13 y = (78 x 10^-3 kg)(9.8)

y = 0.0588 m Or 5.8 cm

so equlibrium position, y0 = 25 - 5.88 = 19.12 cm

hence amplitude , A = 29.7 - 19.12 = 10.58 cm

w = sqrt(k/m) = sqrt(13/0.078) = 12.91 rad/s

y = A cos(12.91t)

for y, that is distance from ground

y\' = A cos12.91t + B

and t = 0 , y= 29.7 cm

29.7 = 10.58 + B

B = 19.12 cm

y = 12.58 cos(12.91t) + 19.12

at t= 0.2s

y = 12.58 cos(12.91x 0.2 rad) + 19.12

y = 8.46 cm

positionn vector, r = <0, 0.0846, 0 > m (from ground) .............Ans

v = dy/dt = d(12.58 cos(12.91t)) /dt = - 162.4 sin(12.91t) cm/s

v = - 1.624 sin(12.91t) m/s

at t = 0.2s

v = 1.624 sin(!2.91 x 0.2) = - 0.862 m/s

momentum = m v = (78 x 10^-3) (-0.862) = - 0.0672 kg m/s

on vector form, <0, - 0.0672, 0 > kg m/s