Energy Conservation and Transfer
Part 1. A 24kg child descends a 5meter(m) high slide and reaches the ground with a speed of 2.8m/s. 1081.92jules(J) of energy was dissiapted due to friction in the process. I found that by using the equation (change in Ug)-K=f. Which (change in Ug) stands for the gravitational potential energy, K stands for the kinetic energy, and f stands for energy lost due to friction. But to use that equation you need to find (change in Ug) and K first. So then you would use (change in Ug)=mgh where m stands for mass(24kg), g stands for the gracitational pull(9.8m/s^2), and h stands for the height(5m). So I plug the numbers in and get (change in Ug)=(24kg)(9.8m/s^2)(5m). That will make (change in Ug) equal to 1176J. Then use the formula, K=(1/2)mV^2, where V^2 is the velocity squared. So, after plugging in those numbers, it is K=(1/2)(24kg)((2.8m/s)^2). That will make K equal to 94.08J. Finally you can plug (change in Ug) and K into the original equation. That would be 1176J-94.08J=f. So, f equals 1081.97J.
The reason why the bar graph of energy is in Ug(the height) and not in column K(kinetic energy) because the man is not moving, therefore, he doesnot have any veloctiy, or in column Us(elastic potential energy because there is no spring attached to the child, or the slide.
The reason why there is Ug and K in the work circle is because they are the only ones that will have any energy at all in the situation. It starts with all the energy in Ug and shifts over to K. That is why there is an arrow pointing from Ug to K. There are three dashes on the arrow because they represent the amount of energy being transfered from Ug to K.
The reason why there is the bar graph of energy just in K is that all the energy has moved to the velocity because there is no spring, so no Us, and is at ground level and not above it, so no Ug
Part 2. A 60kg box is lifted by a rope a distance of 10meters(m) straight up at a constant speed. It requires 1176watts(w) of power to complete this task in 5seconds(s). To get the power of this object you use the equation P=J/t, where P stands for power, J stands for jules, and t stands for time(5s). So, in order to do this equation we need to find J which is all the energy in the object which is only in Ug. But in order to find Ug you use the equation (change in Ug)=mgh, which (change in Ug) is the change in the gravitational potential energy or the height energy, m is the mass(60kg), g is the gravitational pull(9.8m/s^2), and the h is the height of the object(10m). So that is (change in Ug)=(60kg)(9.8m/s^2)(10m), and then (change in Ug)=5880J. Now the first equation can be used. So, P=(5880J)/(5s). That means that P=1176watts of power.
The graphs to the left are the graphs showing the amount of energy that is in the block and rope. K stands for the kinetic energy or the velocity of the object, Ug stands for the gravitational potential energy or the height of the object, and Us stands for elastic potential energy or how much energy the spring has. In the first bar graph they are all at 0 because the object is not moving, is not above the starting location, and there is no spring pushing the object. Then in the second bar graph, K is at 0 because it is not moving, Ug has all of the energy(5880J) because is it above the starting location, and Us is at 0 because there is still no spring in the circuit. And in the Flow circle graph, there is an arrow from the outside pointing to K because there is an outside force making the object move and K has arrow pointing to the Ug because all the energy is being transfer to Ug from K. After the object is done moving, it is above the starting point, so that is why all the energy is in Ug.