00:01
Hi, in the given problem, the length of the rope of the swing is l is equal to 2 .20 meter and the mass of child is m is equal to 25 .0 kilogram.
00:25
The angle, the rope of this swing is making with the vertical at the instant her brother leaves it is given that 40 degree.
00:43
So this theta is 42 degree.
00:48
Now in the first part of the problem, we have to find potential energy of the child at this instant for which.
00:59
Suppose this is the swing oa and here, this is the lower most point of the swing b.
01:12
So in order to find its potential energy at this beginning, we have to find this height h.
01:21
So if we consider this point to be c, we have to find this length b c, this height b c, for which, most of all, we will find this o c .c.
01:32
In the right angle triangle oac hence in right angled triangle oac oc by oa is kosae is cos -theta means cosine of 42 degree so this oc will come out to be oa cause 42 degree or for this oa this is l means 2 .20 meter into cosine of 42 degree, which comes out to be 1 .63 meter.
02:14
So this height bc will become total 2 .20 minus 1 .63 meters.
02:29
Hence, it comes out to be 0 .57 meter.
02:33
So potential energy of the child at point a will be given by ua is equal to mgh or we can say this is 25 .0 kilogram into 9 .8 meter per second square into 0 .57 joules will be the unit of this potential energy.
03:05
So finally this ua here comes out to be 140.
03:10
Jules approximately.
03:12
And here it becomes the answer for this first part of the problem...