A child on a sled is travelling a total distance of 93 m...

A child on a sled is travelling a total distance of 93 m down a dirt hill with an angle of 30 degrees from the horizontal from the top of the hill. The mass of the child and sled is 65 kg. The initial velocity is zero and the final velocity is 5.0 m/s. (a) What is the time it takes to travel down the hill? (b) What is the acceleration down the hill? (c) What is the force of friction? (d) What is the coefficient of kinetic friction?

Transcript

00:01 Hi guys, now in this form a child on a slid is traveling down a distance of 93 meter down a dart hail with the angle of 30 degree from the horizontal from the top of the hill.

00:10 This is the hail we have and there i have this hail is inclined at the angle of 30 degree from this point.

00:23 This angle is 30 degree.

00:28 This length is what 93 meter.

00:33 Now a child is what slid? a child is on a slate and the total distance of 93 meter down a dart hill with an angle of 30 degree from the horizontal from the top of the hill.

00:47 A child is on the slate we have.

00:50 This is child i have.

00:53 It is moving down from the top of the hill.

00:58 This is top of the hill.

01:04 Now when it arrive on the bottom of the hill, the speed of this child is what v is equal to 5 meter.

01:17 Per second and initially u is equal to what is zero when moving downward one is m g acting the components of m g lable will like this and also the friction force will be worked here this is friction force will be worked how can i find out friction force this angle is 30 degree this is sign of i can say sign 30 is but 93 divide by this length of this one which is what this is what this is 93 sign of let me say this is l so l is equal to 93 9c 9x 93 of 93 divided by 10 that is 186 meter let's find out work done due to what friction how can i find out that thing first i need to find out as you can see here if i apply energy conservation energy initial is equal to energy final this one i can say m g.

02:56 This is m g sign of theta and this is m g cos of theta and this is normal which is act like this this is normal i have so normal is equal to m g cos of theta i'm going to say one thing that work done is equal to change in kinetic energy so work done is here first m g h work done due to friction i'm going to say work done due to friction is it just equal to change in kinetic energy or i can say this one will be mgh is equal to mgs minus deta kintechina.

03:52 Walk due to friction.

03:55 And this one i can say m which is 93, g 9 .8, h which is what again, 93.

04:05 M is i guess different.

04:07 M is equal to we have 65...

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