00:01
In the given problem we have a 285 kg stunt boat which is driven on the surface of a lake at a constant speed of 13 .5 meter per second.
00:10
That means the mass of the board given to us is 285 kgs.
00:15
And the constant speed v0 is 13 .5 meter per second.
00:22
And the angle of the ramp is theta is equal to 23 .5 degrees above.
00:30
The horizontal and the coefficient of friction between the board's bottom and the ramp surface given to us that is mu k is 0 .150 now we have the raised height of the ramp let it be h is equal to 1 .8 meters so in part a of the problem we have to assume that the engines are cut off when the boat hits the ramp so what is the speed of the boat as it leaves the ramp let us first look at the free body diagram of the given situation.
01:05
If this is the ramp its height is elevated at 1 .8 meters and this is the boat.
01:14
So the forces acting here are the weight of the boat which is vertically downward denoted as m g that is mass -tens acceleration due to gravity and this is the normal force that is exerted by the ramp on the boat which is denoted by n.
01:31
The frictional force will be in this direction and resolving the force we will get in the opposite direction of n as m g cos theta and here we will get m g sine theta let us assume that the length of this lamp ramp is l so if l is the length of the ramp we will get sine theta is equal to h upon l so the length of the ramp will be h upon sine theta so the height of the ramp is 1 .8 meters and theta is 23 .5 degrees therefore the length of the ramp is 4 .514 meter now let us look at the work done by the friction let it be in the frictional four frictional work done denoted by wf so it is the force times the frictional force times the distance which is the length of the ramp but the formula for the frictional force is mu k n and n is equal to m g cos theta so we will now substitute the value mu k is 0 .150 the mass of the boat is to 85 kg and the acceleration due to gravity is 9 .8 meter per second square times cost of 23 .5 degrees and the length of the ramp that we have seen here is 4 .5 so the work done is 1734 .29 jules.
03:28
Now we will make use of this work done into the conservation of energy theorem.
03:32
So the kinetic energy of the board on the start of the ramp will be equal to, that is half mv square.
03:45
This is the initial kinetic energy will be equal to the kinetic energy of the ramp when it reaches the surface of the water, which is is half m vf square plus the potential energy on reaching the water which is m g h and the frictional work done which is denoted by wf we need to calculate vf so we'll take this term on the left side so we will get half m vf square is equal to half m v i square minus m g h we will now substitute the values.
04:28
So this is half of 285 kg times the initial velocity is 13 .5 meters per second.
04:37
This will be our whole square.
04:39
And the mass year is 285 kgis times the work, the acceleration due to gravity is 9 .8 meter per second and the height is 1 .8 meters.
04:50
Minus the work done by the frictional force is 1 ,734 .29 jules.
04:58
So this is equal to 19 ,208 .935 jules...