Respuesta :
Answer:
a) [tex]\Delta x =0.32433\ m= 324.33\ mm[/tex]
b) [tex]\delta x=0.087996\ m=87.996\ mm[/tex]
c) [tex]\delta x=0.13227\ m=132.27\ mm[/tex]
Explanation:
Given:
- mass of the object, [tex]m=1.4\ kg[/tex]
- height of the object above the spring, [tex]h=1.15\ m[/tex]
- spring constant, [tex]k=300\ N.m^{-1}[/tex]
a)
When the object is dropped onto the spring whole of the gravitational potential energy of the mass is converted into the spring potential energy:
[tex]PE_g=PE_s[/tex]
[tex]m.g.h=\frac{1}{2} \times k.\Delta x^2[/tex]
[tex]1.4\times 9.8\times 1.15=0.5\times 300\times \Delta x^2[/tex]
[tex]\Delta x =0.32433\ m= 324.33\ mm[/tex] is the compression in the spring
b)
When there is a constant air resistance force of 0.6 newton then the apparent weight of the body in the medium will be:
[tex]w'=m.g-0.6[/tex]
[tex]w'=1.4\times 9.8-0.6[/tex]
[tex]w'=1.01\ N[/tex]
Now the associated gravitational potential energy is converted into the spring potential energy:
[tex]PE_g'=PE_s[/tex]
[tex]w'.h=\frac{1}{2} \times k.\delta x^2[/tex]
[tex]1.01\times 1.15=0.5\times 300\times \delta x^2[/tex]
[tex]\delta x=0.087996\ m=87.996\ mm[/tex]
c)
On moon, as per given details:
[tex]m.g'.h=\frac{1}{2} \times k.\delta x^2[/tex]
[tex]1.4\times 1.63\times 1.15=0.5\times 300\times \delta x^2[/tex]
[tex]\delta x=0.13227\ m=132.27\ mm[/tex]
