A mass of 0.3 kg hangs motionless from a vertical spring whose length is 0.77 m and whose unstretched length is 0.59 m. Next the

Question

4 years ago

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A mass of 0.3 kg hangs motionless from a vertical spring whose length is 0.77 m and whose unstretched length is 0.59 m. Next the

mass is pulled down to where the spring has a length of 0.84 m and given an initial speed upwards of 1.8 m/s. What is the maximum length of the spring during the motion that follows?

Physics

2 answers:

katen-ka-za [31] · Answer 1 · 2022-01-10T03:47:11+03:00

Answer:

0.93m

Explanation:

Getting the data:

1. mass (M): 0.3 kg

2. Length with the mass (L1): 0.77 m

3. Unstretched length (Lu): 0.59 m

4. Length after the pulling (Lf): 0.84 m

5. Speed upwards ( $v$ ): 1.8 m/s

By laws of newton and with the first and second data we will find the spring constant using:

M*g = kx

Where g is the gravity, M*g is the weight of the mass, k is the spring constant and x is the stretch (Lu-L1), so:

$k=\frac{M*g}{x}$

$k = \frac{(0.3kg)(9.8m/s^2)}{0.18}$

$k= 16.3$ N/m

Now using the conservation of energy:

$E_1 = E_2$

$\frac{1}{2}Mv^2+\frac{1}{2}kx_1^2=\frac{1}{2}kx_{max}^2$

Where E1 is the energy when the length of the spring is 0.84m, $x_1$ = Lf-Lu, E2 is the energy when the length is maximun and $x_{max}$ is the stretch when the length is maximun. So:

$Mv^2+kx_1^2=kx_{max}^2$

$(0.3kg)(1.8m/s)^2+(16.3N/m)(0.25m)^2=(16.3N/m)x_{max}^2$

$x_{max}=0.34m$

it means that the maximun length of the spring is:

$Lu+x_{max}=0.34m+0.59m=0.93m$

vivado [14] · Answer 2 · 2022-01-10T02:22:11+03:00

Answer:

Maximum spring length = 0.94m

Explanation:

Newton's law of forces in the vertical direction is given as:

Ef = Ks S - mg = 0

Ks = mg/ s

Ks = mg/ (L1 -Lo)

Ks =( 0.3 × 9.8 ) / ( 0.77 - 0.59 )

Ks = 2.94 / 0.18

Ks = 16.33N/m

The initial state is one in which the mass is pulled down so that the length of spring is 0.84m and the mass has initial speed of 1.8m/s. The final state is that in which the minimum point reached and the maximum length attained is L3

Ei = Ef

1/2mVi^2 + 1/2 Ks(L2-Lo)^2 + mg(L3- L2) = 1/2Ks(L3 - Lo)^2.

1/2 Mvi + 1/2 Ks(L2 - Lo)^2 + mg(L3 - L2)^2 = 1/2Ks(L3^2 - 2LoL3 + Lo^2)

1/2 KsL3 - (mg + KsLo)L3 + [1/2KsLo^2 + mgL2 - 1/2Ks(L2 -Lo)^2- 1/2mVi^2] = 0

Substituting the values to solve for L3 using the quadratic formula

1/2(16.33)L3^2 - (0.3×9.8) + 16.33(0.59)L3 + 1/2(16.33×0.59^2) + (0.3×9.8×0.84)- 1/2(16.33(0.84-0.59)^3 - 1/2( 0.3× 1.8^2] = 0

8.165L3^2 - 2.94 + 9.63L3 + 2.84 + 2.47 - 0.51 - 0.486

8.165L3^2 + 9.63L3 + 1.37

L3 =[ 9.63 +- sqrt(9.63^2 - 4(8.165)(1.37) / (2× 8.165)]

L3 =[ 9.63 +- sqrt(92.74 - 44.74) /16.33]

L3 = 9.63 +- sqrt(47.996)/16.43

L3 = 9.63 +- 6.93/ 16.33