00:01
Hi, here in this given problem, mass of the scare that is given as m is equal to 60 .0 kilogram height of the ski slope that is 63 .0 meter.
00:31
In the first part of the problem, work done by the friction when he is coming down.
00:45
That work done is given as wf is equal to minus 10 .9 kilojoules.
00:55
So using work energy theorem half mvf square, final speed of the scare at the bottom of the slope, that will be equal to mgh plus wf.
01:22
So, plugging in all the known values here, this is half times of 60 multiplied by vf square which is missing, we have to find it.
01:36
Mass of the square 60 kilogram into g which is 9 .8 into h which is 63.
01:44
For wf that is minus 10 .9 kilo, means 1000.
01:50
So this vf is calculated to be equal to 29 .5.
01:57
Meter per second speed of the square at the bottom of the slow.
02:03
Okay, now in the second part of the problem, at the soft snow, now at the horizontal, coefficient of kinetic friction mu k, that is 0 .21, and length of this patch, s, this is given as 65 .0 meter.
02:40
Air force means one more type of friction.
02:44
F air, this is given as 180 newton and force of friction fk, this will be given by mu k times normal reaction means weight of the scare...