Apackage of mass 5 kg sits at the equator of an airless asteroid of mass 7.0 1020 kg and radius 3.5 105 m. we want to launch the package in such a way that it will never come back, and when it is very far from the asteroid it will be traveling with speed 165 m/s. we have a large and powerful spring whose stiffness is 1.6 105 n/m. how much must we compress the spring?

3.03m

Explanation:

Assuming that energy is conserved between the initial and the final states.

In the initiat state, from the question, we have both gravitational potential and elastic potential energy, while in the final state, we have only kinetic energy.

Hence:

1/2ks * s^2 + (-GMm/R) = 1/2mvf^2

ks * s^2 = mvf^2 + 2GMm/R

s = √m/ks(vf^2 + 2GM/R)

= √5/1.6*10^5(165^2 + 2(6.67*10^-11)(7.0*10^20)/3.5*10^5

= √(0.00003125 * (27225 + 266800)

= √(0.00003125 * 294025

= √ (9.18828125)

= 3.03m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards