00:01
In this question, our 55 kg square sky is sky is directly down a frictionless slope which is angled at 20 degree to the horizontal.
00:10
Now we have to find fx that is acting on the skyer in the three conditions if the sky velocity is constant increasing at a rate of 1 .1 meter per second or increasing at a rate of 2 .2 meter per second square.
00:28
Right? so first of all, we will draw the figure for this question.
00:32
Now, let us assume this is kair.
00:35
So the forces that would be acting on this object would be the gravitational force vertically downwards, mg.
00:45
And this force mg will have two components.
00:49
One is perpendicular to this normal.
00:56
That would be mg cost theta.
01:00
And one would be along the surface that is m g sine theta there would be an additional force fx as given in the question so if we write down the equation according to the newton's law the net force acting is fx plus m g sine theta and this net force should be equal to mass multiplied by acceleration.
01:39
Now, in the first case where the velocity is constant, the acceleration would be equal to 0.
01:47
Therefore, fx plus mg sine theta is equal to 0...