00:01
In the question given data is, coefficient of static friction mu s equal to 0 .5 and coefficient of kinetic friction mu k equal to 0 .4.
00:23
Velocity of vehicle is 60 mile per hour, which is equal to 26 .8 meter per second.
00:45
We have to find in part a of question, in part a of question, we have to draw the free body diagram of the system.
01:05
In second part of question, we have to find acceleration of rig and acceleration of load with friction.
01:28
In part 3 of the question, we have to find load shift or not, when break is applied.
01:50
In part 4th of the question, we have to find net acceleration.
01:59
In part 6 of the question, we have to find relative velocity of object b with respect to object a.
02:11
First of all, we will find number a.
02:19
Here we will draw the free body diagram of the system.
02:26
So this is load, this is frictional force, this is force due to break, this will be weight of load mg, and it is normal reaction.
03:05
This is free body diagram of the system.
03:12
And part b, we will find acceleration.
03:17
So let us consider acceleration of rig is.
03:24
So change in velocity upon time.
03:26
So velocity is 26 .8.
03:30
Initial velocity of rig is 0 divided by the time is given.
03:35
The question is 5 second.
03:37
So after solving this step, we will get our answer which is equal to 5 .36 meter per second.
03:49
Square along x -axis.
03:55
This is acceleration of rig.
04:07
Now in the same part we have to find acceleration of load.
04:14
Let us consider acceleration of load is a load and it will be equal to minus mu s g using concept of frictional force and and newton's second law, we will obtain this formula, a load equal to minus mu sg.
04:43
Now, after putting the value, it is 0 .5 multiplied by 9 .8...