Use the approximation that for each time step A spring with

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?

Solution

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