Question

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?

Answer 1

1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

$F=mg=(7.6 kg)(9.8 m/s^2)=74.5 N$

And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

$F=kx\\k=\frac{F}{x}=\frac{74.5 N}{0.29 m}=256.9 N/m$

2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s$

And the frequency of oscillation is given by:

$\omega=2\pi f\\f=\frac{2 \pi}{\omega}=\frac{2\pi}{5.81 rad/s}=1.08 Hz$

3) 2.19 m/s

The velocity at time t of the block is given by:

$v=v_0 cos (\omega t)$

where

$v_0 = 4.4 m/s$ is the initial velocity of the block

$\omega=5.81 rad/s$ is the angular frequency

t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

$v(0.36 s)=(4.4 m/s) ( cos ((5.81 rad/s)(0.36 s)) = -2.18 m/s$

And the negative sign means that the direction of the velocity is upward (because the initial velocity was downward)

4) 25.6 m/s^2

The maximum acceleration is given by:

$a_0 = \omega^2 A$

where A is the amplitude of the oscillation.

We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

$K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m$

So, the maximum acceleration is

$a_0 = \omega^2 A=(5.81 rad/s)^2 (0.76 m)=25.6 m/s^2$

5) 95.1 N

The magnitude of the net force acting on the block is given by the difference between the weight and the restoring force of the spring:

$F=mg-kx$

First, we need to find the position x at t=0.36 s, which is given by

$x(t)=A sin(\omega t)=(0.76 m)(sin ((5.81 rad/s)(0.36 s))=0.66 m$

And so, the net force is

$F=(7.6 kg)(9.8 m/s^2)-(256.9 N/m)(0.66 m)=-95.1 N$

And the negative sign means the direction of the force is upward.

Answer 2

Answer:

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