9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T
hen a second spring with equilibrium length L2 and spring constant k2 is hung from the bottom of m1. Mass m2 is suspended from this second spring. How far is m2 below the ceiling
1 answer:
Answer:
The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.
See attachment below for full solution
Explanation:
This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.
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Answer:
F = 4.48N
Explanation:
In order to calculate the net gravitational force on the rocket, you take into account the formula for the gravitational force between two objects, which is given by:
(1)
G: Cavendish's constant = 6.674*10^-11 m^3kg^-1s^-2
r: distance between the objects
You have a rocket at the middle of the distance between Earth and Moon, then, you have opposite forces on the rocket.
If you assume the origin of a system of coordinates at the rocket position, with the Moon to the left and the Earth to the right, you have:
(2)
Me: mass of the Earth = 5.98*10^24 kg
Mm: mass of the Moon = 7.35*10^22 kg
m: mass of the rocket = 1200kg
r1: distance from the rocket to the Earth = 3.0*10^8m
r: distance between rocket and Moon = 3.84*10^8m - 3.0*10^8m = 8.4*10^7m
You replace the values of the parameters in the equation (2):
The net force exerted over the rocket is 4.48N