Question

A small ball with a mass of 1 kg rolls down a long frictionless inclined ramp, which is at an angle 3o Degree
above the horizon. A linear spring, whose length is not negligible, is attached to the bottom of the
ramp. The ball is released from a distance of 1 m above the spring, measured along the plane and
the spring constant is 1N/m, then the spring is depressed by

Answer 1

So what is the question ?

Answer 2

The spring is depressed by how much? The answer is 1.6m. But how?

Answer 3

ball rolls distance 1 + x
loss of height = (1+x)sin 30 = .5(1+x)
loss of potential energy = 1*9.8*.5(1+x)

gain of potential energy = .5 k x^2 = .5 x^2
so
9.8(1+x) = x^2

x^2 - 9.8 x - 9.8 = 0

x = [ 9.8 +/-sqrt(96+39.2)]/2

x = (9.8+/-11.6)/2

x = 10.7

Answer 4

The answer is the spring is depressed by 1.6m

Answer 5

Using the mass of the air (4.65MJ). We can assume that the mass of the clouds (which is made of air) is directly proportional. With 4.65 Micheal Jacksons, the amount of Micheal Jordans is directly equivalent. MJ = MJ. We know that Jackson, Tyson, Jordan Game 6 are lyrics to the song "What is Love" we can assume that the force used is equivalent to Haddeways belt mass: 666 N. This number is the illuminati's second elites squads badge ID, therefore the spring is depressed by 1.6m