Respuesta :
Answer:
Explanation:
spring constant of spring = mg / x
= .4 x 9.8 / ( .95 - .65 )
=13.07 N / m
energy stored in spring = 1/2 k x²
= .5 x 13.07 x ( 1.2 - .65 )²
= 1.976 J
Let it goes x m beyond its equilibrium position
Total energy at initial point
= 1.976 + 1/2 m v²
= 1.976 + .5 x .4 x 1.6²
= 2.488 J
energy at final point
= mgh + 1/2 k x²
.4 x 9.8 x ( .55 + x ) + .5 x 13.07 x² = 2.488
6.535 x² + 2.156 + 3.92 x = 2.488
6.535 x² + 3.92 x - .332 = 0
x = .075 m
7.5 cm
The maximum length of the spring is 1.4m during the motion.
Spring-mass system:
The stretch or displacement of the spring is:
x = 0.95m - 0.65m
x = 0.30m
in equilibrium position:
mass of object = restoring force of spring
mg = kx
where k is the spring constant
k = mg/x
k = (0.4×9.8)/0.30
k ≈ 13 N/m
when the spring is stretched to 1.2 m, the potential energy stored in the spring:
[tex]PE_i=\frac{1}{2}kx^2=\frac{1}{2}\times13\times(1.2-0.65)^2\\\\PE_i=1.966\;J[/tex]
The object is given an initial speed v = 1.6 m/s, then the initial kinetic energy:
[tex]KE_i=\frac{1}{2}\times0.4\times(1.6)^2\\\\KE_i=0.512\;J[/tex]
let the maximum stretch of the spring be x', at this point the velocity will be 0, so only potential energy will be present, say [tex]PE_f[/tex]
if we consider the lowest point as ground level, h = 0, then the maximum height obtained will be
= mg(1.2-0.65+x)
= mg(0.55 + x)
so, [tex]PE_f=\frac{1}{2}kx'^2+mg(0.55+x)[/tex]
Then, from conservation of energy:
[tex]KE_i+PE_i=PE_f\\\\0.512+1.966=\frac{1}{2}\times13x'^2+0.4\times9.8(0.55+x)\\\\6.535 x^2 + 3.92 x - 0.332 = 0\\\\x'=0.75m[/tex]
So the maximum stretch is 0.75m.
Therefore the maximum length of the spring will be = 0.65+0.75 = 1.4m
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