Respuesta :
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
L3 = (9.63 - 6.93)/16.34
L3 = 2.7/16.33
L3 = 0.165 + 0.77
L3= 0.935m
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 ([tex]v[/tex]): 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:
[tex]k=\frac{M*g}{x}[/tex]
[tex]k = \frac{(0.3kg)(9.8m/s^2)}{0.18}[/tex]
[tex]k= 16.3[/tex] N/m
Now using the conservation of energy:
[tex]E_1 = E_2[/tex]
[tex]\frac{1}{2}Mv^2+\frac{1}{2}kx_1^2=\frac{1}{2}kx_{max}^2[/tex]
Where E1 is the energy when the length of the spring is 0.84m, [tex]x_1[/tex] = Lf-Lu, E2 is the energy when the length is maximun and [tex]x_{max}[/tex] is the stretch when the length is maximun. So:
[tex]Mv^2+kx_1^2=kx_{max}^2[/tex]
[tex](0.3kg)(1.8m/s)^2+(16.3N/m)(0.25m)^2=(16.3N/m)x_{max}^2[/tex]
[tex]x_{max}=0.34m[/tex]
it means that the maximun length of the spring is:
[tex]Lu+x_{max}=0.34m+0.59m=0.93m[/tex]
