00:01
To the kinematics in which we have a block of mass m equals to 15 kg which is over an inclined plane of theta equals to 50 degree regards to the horizontal one so the diagram for this question can be drawn like this okay so this will be the diagram for this situation this angle is theta which is equals to 50 degree and this is the block as per the question and this is the weight and these are the component of the weight this is m g sine theta and mg cos -teta will be in this direction.
00:32
M -g -cost -theta.
00:35
And this will be the normal force and this will be the friction force.
00:39
And we have coefficient of friction mu that is equal to 0 .12 and length of the plane l that is equal to 20 meter.
00:48
So for the part 1 of the question, we have to calculate the acceleration of the falling block.
00:54
So from the newton's law in the inclined plane, so we can write that mass into acceleration, this is equals to m g, sine theta minus mu mg, cost theta.
01:08
So from here we get acceleration a equals to g bracket, sine theta minus mu cost theta.
01:16
So substituting values, so we get acceleration a equals to g which is 9 .80 and sine theta is 50 degree minus 0 .12 and cos 50 degree.
01:29
So from here after solving, we get acceleration a that is equals to 6 .75 meter per second square.
01:37
So this become the answer for the part one of the question.
01:41
Now, moving to the part two in which we have to calculate if the block start at rest, that is initial speed is equals to 0, then what is the final speed when it reaches the ground? when it reaches at this point after traveling this distance l.
01:57
So, from the equation of kinematics, we can write that v square equals to u square plus 2a l...