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

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

Answer 2

The correct responses (approximate values) are;

1) 257.09 N/m

2) 0.93 Hz

3) 2.2 m/s upwards

4) 25.6 m/s²

5) 168.49 N

Methods used to derive the above responses

Given:

Mass of the block, m = 7.6 kg

Extension of the spring x = 0.29 m

Velocity with which the block is pushed, v = 4.4 m/s

1) Applied force due to weight of the block, F = m × g

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Force, F = k·x

Where;

k = The spring constant

Therefore;

$k = \dfrac{F}{x} = \dfrac{m \cdot g}{x}$

Which gives;

$k = \dfrac{7.6 \times 9.81 }{0.29} \approx \mathbf{257.0896552}$

• The spring constant, k = 257.0896552 N/m (which is approximately 257.09 N/m)

2) The frequency of oscillation is given by the formula;

$f = \mathbf{\dfrac{1}{2 \cdot \pi} \cdot \sqrt{\dfrac{k}{m} } }$

Therefore;

$f = \dfrac{1}{2 \times \pi} \times \sqrt{\dfrac{257.0896552}{7.6} } \approx \mathbf{0.9256688223}$

• The frequency of oscillation, f ≈ 0.9256688223 Hz (approximately 0.93 Hz

3) Potential energy of the spring, $P.E. = \frac{1}{2} \cdot k \cdot x^2$

Kinetic energy of the, $K.E = \frac{1}{2} \cdot m \cdot v^2$

At the maximum extension of the spring, we have;

P.E. = K.E.

Which gives;

$\mathbf{\dfrac{1}{2} \cdot k \cdot ((\Delta x)_{max})^2} = \dfrac{1}{2} \cdot m \cdot v^2$

$(\Delta x)_{max} = \mathbf{ \sqrt{\dfrac{m \cdot v^2}{k} } }$

Therefore;

$(\Delta x)_{max} = \sqrt{\dfrac{7.6 \times 4.4^2}{257.0896552} } \approx 0.75651435$

The equation of motion of a spring is x = A·sin(ω·t + ∅)

Where;

Initial stretch of the spring = 0.29 m

A = $(\Delta x)_{max}$ = 0.75651435

ω = 2·π × f

∅ = 0

We have;

Δx = 0.29 + 0.75651435·sin(2·π·0.9256688223·t)

$v(t) = \dfrac{\Delta x}{\Delta t} = 0.75651435 \times cos(2\cdot \pi \cdot 0.9256688223 \cdot t) \times (2\cdot \pi \cdot 0.9256688223)$

At t = 0.36 s, we have;

$v(t) = 0.75651435 \times cos(2 \cdot \pi \cdot 0.9256688223 \times 0.36) \times (2 \cdot \pi \cdot 0.9256688223) = -2.1977836$

• The velocity at 0.36 s ≈ 2.1977836 m/s in the upward direction (approximately 2.2 m/s)

4) The magnitude of the maximum acceleration is found as follows

$The \ applied \ force, \ F = \mathbf{k \times (\Delta x)_{max}}$

Which gives;

$F$ = 257.0896552 N/m × 0.75651435 m

F = m·a

Where;

a = The acceleration

Therefore;

$a_{max} = \dfrac{F}{m} = \mathbf{ \dfrac{k \times (\Delta x)_{max}}{m} }$

Which gives;

$a_{max} = \dfrac{257.0896552 \, N/m \times 0.75651435 \, m}{7.6 \, kg} \approx 25.591054 \, m/s^2$

• The maximum acceleration, $a_{max}$ ≈ 25.591054 m/s² (approximately 25.6 m/s²)

5) The net extension of the spring at 0.36 s is given as follows;

Δx(0.36) = 0.75651435 × sin(2·π×0.925668823×0.36)

F = k × x

Therefore;

F(0.36) = 257.0896552 × (0.75651435 × sin(2·π×0.925668823 × 0.36) = 168.49

• The net force at t = 0.36 s is 168.49 N

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