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A stationary mass m = 1.3 kg is hanging from a spring of spring constant k = 1200 n/m. you raise the mass a distance of 10 cm above its equilibrium position. how much has the potential energy of the mass-spring system changed?

Respuesta :

skyluke89
The weight of the mass is
[tex]F=mg=(1.3 kg)(9.81 m/s^2)=12.8 N[/tex]
This force stretches the spring of a certain amount [tex]x_1[/tex], which can be calculated by using Hook's law:
[tex]x_1 = \frac{F}{k} = \frac{12.8 N}{1200 N/m} =0.011 m[/tex]
with respect to the equilibrium position of the spring.
So, the initial elastic potential energy of the spring is
[tex]U_1 = \frac{1}{2} kx_1^2 = \frac{1}{2} (1200 N/m)(0.011 m)^2=0.07 J[/tex]

Later, the spring is compressed by [tex]x_2 = -10 cm =-0.100 m[/tex] from its equilibrium position (the negative sign means that now it is a compression, while before it was an elongation). This means that the new elastic potential energy of the spring is
[tex]U_2 = \frac{1}{2} kx_2^2 = \frac{1}{2}(1200 N/m)(-0.100 m)^2 = 6 J [/tex]

So, the change in elastic potential energy of the mass-spring system is
[tex]\Delta U=U_2 -U_1 = 6J-0.07 J=5.93 J[/tex]

However, if we consider also the change in gravitational potential energy of the mass, the answer is different. In fact, the height of the mass changed by 
[tex]\Delta h=x_1 -x_2 = 0.011 m-(-0.100 m)=0.111 m[/tex]
So its gravitational potential energy increased by
[tex]\Delta E_{gp} = mg \Delta h=(1.3 kg)(9.81 m/s^2)(0.111 m)=1.42 J[/tex]

So, the total change of potential energy (elastic+gravitational) of the mass-spring system is
[tex]\Delta U=5.93 J+1.42 J=7.35 J[/tex]
onyebuchinnaji

The change in the potential energy of the mass-spring system is 7.27 J.

The given parameters;

  • mass, m = 1.3 kg
  • spring constant, k = 1200 N/m
  • height the mass is raised, h = 10 cm = 0.1 m

The change in the potential energy of the mass-spring system is calculated as follows;

total change in potential energy =  change in potential energy due to height of the mass  +  change in potential energy due to compression of the spring.

[tex]\Delta P.E _{total} = mg\Delta h + \frac{1}{2} kx^2\\\\\Delta P.E _{total} = (1.3 \times 9.8 \times 0.1) \ + \ (\frac{1}{2} \times 1200 \times 0.1^2)\\\\\Delta P.E _{total} = 7.27 \ J[/tex]

Thus, the change in the potential energy of the mass-spring system is 7.27 J.

Learn more here:https://brainly.com/question/20388610

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