00:01
Problem 740, a 2 kilogram block is pushed against a spring with an actual mass and force constant k equals 400 newtons per meter and it compresses 0 .22 meters.
00:12
When a block is released, it moves along a frictionless horizontal surface and then up the frictionless incline with the slope 37 degrees.
00:20
That's not really to the best scale.
00:22
What does the speed of the block has a slides along a horizontal surface after having left the spring? so part a.
00:27
So part a, it's a straightforward, conservation of energy problem we have we have the potential elastic energy is going to be equal to the kinetic energy of the block after it is released from the spring so to do that it's just going to be one and a half k x squared is equal to one half m v squared and then we just need to substitute everything so our mass is given as two kilograms k is given as nons per meter, x is the distance compressed, and we're solving for v.
01:09
So if we plug in all those values, times 400 times 0 .22 meters squared, equal to one half times two kilograms, velocity squared, we get v is equal to 3 .11 meters per second.
01:42
Okay, then for b, how far does a block travel up the incline before starting to slide back down.
01:49
So what we can do is we can again write what the conservation energy between the block in position as it is over here versus over here.
01:59
And the difference here is that it has kinetic energy and no potential energy.
02:03
If we set this to be y equals zero or potential energy equals zero, gravity function energy equals zero.
02:11
And up here it has stopped before it starts to slide back down.
02:15
So it's connected energy zero.
02:17
So we can do is we can say, equal to kinetic energy initial...