00:01
Hi there, here you have to solve the following problem.
00:03
So sam worked on developing power by pushing a football sled during his break and lunch.
00:12
So here we assume that he applied a horizontal force of 200 newtons, directed to the right for two and a half seconds on a sled which has a mass of 40 kilograms.
00:30
And initially at rest.
00:32
The coefficient of friction between the sled and the tarf is 0 .35 and first we have to calculate the momentum of the sled after 2 .5 seconds.
00:55
Let's do this.
00:56
So let's illustrate the system.
00:58
So this is a ground.
01:00
This is a sled and sam is applying force f to the right.
01:04
There is a kinetic friction.
01:06
To the left, reaction is upwards, and gravity is downwards.
01:12
Let's introduce y and x -axis.
01:14
And if we project the second newton's low equation on the x -axis, f minus kinetic friction equals to m -a, and that is f minus m -k times m -g.
01:30
Now let's rewrite the second -newton's low equation in terms of momentum of the force.
01:36
So, f minus mu k times m g, f minus mu k m g times delta t equals to change of the momentum, which is mv final minus mv initial.
01:54
So the goal is accelerated from the rest, therefore that just equals to p, which we have to calculate.
02:06
Then this p, momentum, equals to 200 newtons minus 0 .35 times 9 .80 meters per second squared times 40 kilograms.
02:23
And this is all multiplied by 2 .5 seconds.
02:32
Let's complete this calculation.
02:55
That equals to 157 kilogram meter per second...