00:01
This question we have to use the work energy through them to solve each problem so we can use newton loss to check.
00:06
So skier is moving at 5 meter per second and encounters a long, rough horizontal patch of snow having a coefficient of friction as 0 .22 with her skis.
00:20
How far does she travel on this patch before stopping? so this is where he's traveling at 5 meter per second and this is a rough pack.
00:33
Having coefficient of kinetic friction at 0 .22 and it reaches a point and it stops.
00:39
So as for the as for the work energy theorem, the initial kinetic energy is actually all converted into the work done by the friction because after all it stops.
00:50
So it's energy, there is no kinetic and there is no potential because why put in no potential because it is at the same surface.
00:56
So initial kinetic is half m5 square and the work done by the friction is the force of friction times the displacement, which is something which we need to find.
01:08
So this becomes half m 25 and the force of friction is over here.
01:16
Normal reaction is here and this is the weight and this is the motion.
01:19
So force of friction is mu k times normal reaction times d.
01:24
So this becomes 25m over 2 is equal to mu k, which is 0 .22 times the normal reaction which is weight, which is nothing but m .g.
01:34
So m is not given.
01:36
It's not even required and g is over here.
01:38
So this m and m is cancelled.
01:40
And if we rearrange this a little bit, the value of d comes out as 25 over 2 times 0 .22g.
01:49
So those are a required value of the distance which it travels.
01:56
So 25 over 2 over 0 .22 over 9 .8 is coming as 5 .7.
02:03
In fact 5 .80 meters of two decimal places.
02:08
So this the distance which it travels.
02:12
Now part b is suppose the rough patch and part is only for 2 .9 meter long.
02:19
So offers the skiers moving when she reached the end of the patch.
02:23
So over here we'll have that is just for 2 .9 meters and the solution is going to look like over here will have mass traveling with the velocity of v then there's a patch of 2 .9 meters over here so we have some speed over here as well let's call it v2 and the initial velocity was already given to us as 5 meter per second so this is 5 meter per second so using the energy theorem the initial kinetic energy is converted into the work done by the friction plus the final kinetic energy.
03:09
So the initial kinetic energy is half m5 square...