00:01
Here in part a, the diagram is as follows.
00:05
Here, there are two blocks, block a and block b.
00:09
The distance to block a is 6 .6 meter and there is an angle 20 degree here.
00:14
The frictional force f is equal to coefficient of friction into normal reaction and this is equal to coefficient of friction into normal reaction can be written as mg cos theta.
00:27
Here the coefficient of friction is given by fk is equal to 0 .2 plus 0 .15, so it sums up to 0 .35.
00:42
The mass of the system is equal to 5 kilogram plus 10 kilogram and this sums up to 15 kilograms.
00:52
Now, by newton's second low, we know that the equation for force is given by force in x direction will be equal to mass into the acceleration in the direction.
01:09
Or this can be written as here, force due to gravity in x direction minus force due to kinetic friction will be equal to mass into acceleration in x direction.
01:20
Or this can be written as fgx can be.
01:23
Be expressed as m g into sine theta minus f can be represented as mu into m g into cos theta is equal to mass into acceleration a x or from here acceleration a x can be represented as m g sine theta minus mu into m g g cos theta divided by m m is a common term so it gets cancelled so here, x will be equal to g sine theta minus mu into g cos theta.
02:01
Upon substituting the values, g is 9 .81 meter per second square into sine.
02:07
Here value of theta is 20 degree, which is the inclination, minus.
02:12
Here the value of mu is 0 .35 into 9 .81 into cos 20 degree.
02:20
Upon solving we will get a x is equal to 0 .13 meter per second square.
02:29
Now by using equation of motion, we know that displacement s is equal to initial velocity u into time taken t plus half into acceleration a into time square.
02:42
Here this term becomes zero because initial velocity is zero.
02:46
So displacement s will be equal to half into acceleration into time square.
02:50
Or from this equation we can write an expression for time.
02:54
Time will be equal to square root of two times the displacement divided by acceleration a.
03:00
Here, square root of two times, here the displacement is actually the distance d, which is 6 .6 meter divided by acceleration.
03:09
We just now found out which is 0 .13 meter per second square.
03:13
On solving, we will get time taken to be 10 .08 seconds.
03:17
So this is the time taken for the package a in order to reach the bottom.
03:27
So this will be the answer of part a.
03:32
So 10 .080 seconds will be the answer...