Question

0.35 m. When 45 N/m and relaxed length ao 1) A bob of mass m is suspended by a spring with spring constant k the bob and the spring are located on the surface of Planet X as shown, the equilibrium length eg,X of the spring is = 0.55 m. The mass and the radius of Planet X are Mx 2x 10 kg and Rx 5 x 10° m, respectively. Ieqx Xeqx m Rx Mx Bob and spring on Planet X. Note, the sizes of the spring and bob are greatly exaggerated.

Answer 1

given
k = 45 N/m
delta_x = 0.55 - 0.35 = 0.20 m
Mx = 2*10^24 kg
Rx = 5*10^6 m

I think we need to find m here.

acceleration due to gravity on the planet,

g = G*Mx/Rx^2

= 6.67*10^-11*2*10^24/(5*10^6)^2

= 5.336 m/s^2

in the equilibrium, Fg = F_spring

m*g = k*delta_x

m = k*delta_x/g

= 45*0.2/5.336

= 1.69 kg <<<<<<<---------------Answer

Please comment for further help on this problem.