Respuesta :
Answer:
a) If the mass get double then the potential of the spring gets four times.
b) P'=3.848 J
Explanation:
Given that
P= 0.962 J
m = 3.5 kg
m'= 7 kg
Lets take extension in the spring is x when the mass 3.5 kg is attached to the spring.
m g = K x
K=Spring constant
[tex]x=\dfrac{mg}{K}[/tex]
We know that potential energy given as
[tex]P=\dfrac{1}{2}Kx^2[/tex]
[tex]P=\dfrac{1}{2}K\times \dfrac{m^2g^2}{K^2}[/tex]
[tex]P=\dfrac{1}{2}\times \dfrac{m^2g^2}{K}[/tex]
If the mass get double then the potential of the spring gets four times.
[tex]P'=\dfrac{1}{2}\times \dfrac{(2m)^2g^2}{K}[/tex]
[tex]P'=4\times \dfrac{1}{2}\times \dfrac{m^2g^2}{K}[/tex]
P'= 4 P
When mass ,m' = 7 kg
Then potential will be
[tex]0.962=\dfrac{1}{2}\times \dfrac{3.5^2\times g^2}{K} [/tex] -----1
[tex]P'=\dfrac{1}{2}\times \dfrac{7^2\times g^2}{K}[/tex] -------2
From equation 1 and 2
[tex]\dfrac{0.962}{P'}=\dfrac{\dfrac{1}{2}\times \dfrac{3.5^2\times g^2}{K} }{\dfrac{1}{2}\times \dfrac{7^2\times g^2}{K} }[/tex]
P'= 4 x 0.962 J
P'=3.848 J
Answer:
[tex]k=611.517\ N.m^{-1}[/tex]
[tex]U_7=3.8477\ J[/tex]
Explanation:
Given:
- spring potential energy stored due to hanging mass, [tex]U=0.962\ J[/tex]
- mass attached to the spring, [tex]m=3.5\ kg[/tex]
Now the force on the mass due to gravity:
[tex]F=m.g[/tex]
[tex]F=3.5 \times 9.8[/tex]
[tex]F=34.3\ N[/tex]
This force pulls the spring down, so:
[tex]F=k.\delta x[/tex]
[tex]34.3=k\times \delta x[/tex] ....................(1)
For the spring potential:
[tex]U=\frac{1}{2} k.\delta x^2[/tex]
[tex]0.962=0.5\times k\times \delta x^2[/tex]
[tex]1.924=k\times \delta x^2[/tex] .........................(2)
Using eq. (1) & (2)
[tex]\frac{1.924}{x^2} =\frac{34.3}{x}[/tex]
[tex]x=0.05609\ m[/tex]
a.
Now the spring factor:
using eq. (1)
[tex]34.3=k\times \delta 0.05609[/tex]
[tex]k=611.517\ N.m^{-1}[/tex]
b.
when mass attached is 7 kg.
The spring potential energy:
[tex]U_7=\frac{1}{2} \times k.\delta x'^2[/tex] ............(3)
Now the force on the mass due to gravity:
[tex]F=m.g[/tex]
[tex]F=7\times 9.8[/tex]
[tex]F=68.6\ N[/tex]
This force pulls the spring down, so:
[tex]F=k.\delta x[/tex]
[tex]68.6=611.517\times \delta x[/tex]
[tex]x=0.11218\ m[/tex]
Using eq. (3)
[tex]U_7=\frac{1}{2}\times 611.517\times 0.11218^2[/tex]
[tex]U_7=3.8478\ J[/tex]
