Respuesta :
Answer:
0.011 m.
Explanation:
Energy stored in the spring = Energy of the projectile.
1/2ke² = mgh ................ Equation 1
Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.
make e the subject of the equation
e = √(2mgh/k)............................. Equation 2
Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m
Constant: g = 9.8 m/s²
Substitute into equation 2
e = √(2×0.015×5/1200)
e = √(0.15/1200)
e = √(0.000125)
e = 0.011 m.
Answer:
The spring compressed initially [tex]3.5 \ cm[/tex]
Explanation:
Spring gun's force Constant [tex]$k=12 \mathrm{~N} / \mathrm{cm}$[/tex]
Gun's verticle projection [tex]=15-g[/tex]
Height [tex]=5.0 \ m[/tex]
Initial Compression[tex]=?[/tex]
Step 1:
The formula for elastic potential energy of the system.
[tex]$E_{e}=\frac{1}{2} k d^{2}$[/tex]
Here, [tex]$E_{e}$[/tex] is the elastic potential energy of the spring, [tex]$k$[/tex] is the spring constant and [tex]$d$[/tex] is the extension of the spring.
The energy stored in the compressed spring is elastic potential energy. The elastic potential energy of the spring gets converted to the gravitational potential energy of the projectile.
Step 2:
The equation for the gravitational potential energy of the projectile:
[tex]$E_{p}=m g h$[/tex]
Here, [tex]$E_{p}$[/tex] is the Gravitational potential energy, [tex]$m$[/tex] is the mass of the projectile, [tex]$g$[/tex] is the acceleration due to gravity and [tex]$h$[/tex] is the height reached by a projectile.
Convert spring constant [tex]$k$[/tex] from [tex]$\mathrm{N} / \mathrm{cm}$[/tex] to [tex]$\mathrm{N} / \mathrm{m}$[/tex].
[tex]$k=(12 \mathrm{~N} / \mathrm{cm})\left(\frac{10^{2} \mathrm{~N} / \mathrm{m}}{1 \mathrm{~N} / \mathrm{cm}}\right)$[/tex]
[tex]$=1200 \mathrm{~N} / \mathrm{m}$[/tex]
Convert mass from [tex]$g$[/tex] to [tex]\mathrm{kg}$[/tex].
[tex]$m=(15 \mathrm{~g})\left(\frac{10^{-3} \mathrm{~kg}}{1 \mathrm{~g}}\right)$[/tex]
[tex]$=0.015 \mathrm{~kg}$[/tex]
Step 3:
From the law of conservation of energy, elastic potential energy converts to gravitational potential energy. Therefore,
[tex]$E_{e}=E_{p}$[/tex]
Substitute [tex]$\frac{1}{2} k d^{2}$[/tex] for [tex]$E_{e}$[/tex] and [tex]$m g h$[/tex] for [tex]$E_{p}$[/tex]
[tex]$\frac{1}{2} k d^{2}=m g h$[/tex]
Rearrange the equation in terms of [tex]$d$[/tex]
[tex]$d=\sqrt{\frac{2 m g h}{k}}$[/tex]
Step 4:
Substitute [tex]$0.015 \mathrm{~kg}$[/tex] for [tex]$m[/tex],
[tex]9.81 \mathrm{~m} / \mathrm{s}^{2}$[/tex] for [tex]$g[/tex],
[tex]5 \mathrm{~m}$[/tex] for [tex]$h$[/tex] And
[tex]$1200 \mathrm{~N} / \mathrm{m}$[/tex] for [tex]$h$[/tex]
[tex]$d=\sqrt{\frac{2(0.015 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5 \mathrm{~m})}{1200 \mathrm{~N} / \mathrm{m}}}$[/tex]
[tex]&=0.035 \mathrm{~m} \\[/tex]
[tex]&=0.035 \mathrm{~m}\left(\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\right) \\[/tex]
[tex]&=3.5 \mathrm{~cm} \end{aligned}$[/tex]
Thus we can say the initial compression of the spring is [tex]$3.5 \mathrm{~cm}$[/tex]
To learn more about massless spring, refer:
- https://brainly.com/question/14237885
- https://brainly.com/question/15498586
