Block with mass m =7.6 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.29 m. while at this equilibrium position, the mass is then given an initial push downward at v = 4.4 m/s. the block oscillates on the spring without friction. 1) what is the spring constant of the spring? 257.0896552 n/m submit 2) what is the oscillation frequency? 0.9256688223 hz submit 3) after t = 0.36 s what is the speed of the block? 2.197783607 m/s submit your submissions: 2.197783607 computed value: 2.197783607submitted: wednesday, november 1 at 7: 15 pm feedback: correct! 4) what is the magnitude of the maximum acceleration of the block? 25.59105447 m/s2 submit your submissions: 25.59105447 computed value: 25.59105447submitted: wednesday, november 1 at 7: 27 pm feedback: correct! 5) at t = 0.36 s what is the magnitude of the net force on the block?

a.257.0896552

b. 0.9256688223 hz

c.2.197783607 m/s

d.25.59105m/s2

e.205.66 N

Explanation:

Block with mass m =7.6 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.29 m. while at this equilibrium position, the mass is then given an initial push downward at v = 4.4 m/s. the block oscillates on the spring without friction. 1) what is the spring constant of the spring? 257.0896552 n/m submit 2) what is the oscillation frequency? 0.9256688223 hz submit 3) after t = 0.36 s what is the speed of the block? 2.197783607 m/s submit your submissions: 2.197783607 computed value:2.197783607submitted:wednesday, november 1 at 7:15 pm feedback:correct! 4) what is the magnitude of the maximum acceleration of the block? 25.59105447 m/s2 submit your submissions: 25.59105447 computed value:25.59105447submitted:wednesday, november 1 at 7:27 pm feedback:correct! 5) at t = 0.36 s what is the magnitude of the net force on the block?

force exerted on an elastic material is directly proportional to te extension provided the elastic limit is not exceeded , by hookes' law

F = kx = m*g = k*.2

k =  m*g/.2 = 7.6*9.81/.29 = 257.089n/m

The frequency of oscillation is:

f = sqrt( k/m ) / ( 2*π ) = sqrt( 257.089 / 7.6 ) / ( 2*π ) =  0.92566Hz

The kinetic energy at t = 0 is:

E = (1/2)*m*v^2 = (1/2)*7.6*(4.4)^2 = 73.56 J

potential energy at wen the acceleration becomes maximum at th e edge

Ep = (1/2)*k*(Δx)^2 = E

Δx = sqrt( 2*E / k ) = sqrt( 2*73.56 / 257.089 ) = 0.75 m

The additional force of the spring is:

F = k*Δx = 257.089*0.75 = 192.81 N

F = m*a

a = F/m = 192.81/7.6=25.37 m/s^2  

The equation of motion of the block

x = .29 + .75*Sin( 2*π*0.92*t)

Choose the Sin term for the motion, since the additional displacement is zero at t = 0.

The speed of the block is:

v(t) = dx/dt =.75*[ Cos( 2*π*0.92*t ) ]*(2*π*0.92)

at time .36sec

v(.36) = .75*5.780*Cos( 2.08) = -2.16 m/s

This means that the mass is moving upward at 2.16 m/s.  

According to the equation of motion, the x displacement at 0.36 s is:

x(.36) = 0.2  + .75*Sin(2*π*0.92 *.36 ) = 0.80 m

recall th at f=kx

F = k*x = 257.08* ( .8 ) = 205.66 N