Energy!!!!!

Energy: Worksheet 5

1. A 60kg box is lifted by a rope a distance of 10 meters straight up at a constant speed. How much power is required to complete this task in 5 seconds?

To begin, we must draw our initial energy bar diagram. Since we are assuming that the box is stationary before it is lifted, there is no K value. This is because the K value represents kinetic energy, meaning the box must have velocity in order to have a K value, which it does not since it is not initially moving. Also, we are assuming that the box is being lifted off the ground - since it is not in the air, it has no Ug. Ug is gravitational potential energy, meaning the object must have height above a surface so it will have potential energy to expend if it falls. Since the box is on the ground, it has no height above the ground, meaning it has no Ug value. Lastly, there is no spring involved in our system, so therefore there is no Us. Us represents spring potential energy, which involves the compression or stretching of a spring. Since we did not include a spring in our system, there is no Us value present.


We then must figure out where the energy is at the end, as well as how it gets there, so we will make an energy flow diagram and final energy bar diagram. (Hint: It is best to make the bar diagram first so you can account for that energy in the flow diagram.)To include the three bars of Ug that we had in the final energy bar diagram shown below, we must add an outside force to the flow diagram since we did not have any energy in our initial energy bar diagram. This outside force is tension in this case since a rope connected to the box is being pulled, therefore raising the box upward. We include three bars of energy on the arrow going from outside the circle to inside. That energy went to kinetic since the box was moving. Then, once it finished moving, all three bars had been transferred to Ug since it was then in the air and had potential gravitational energy.At the end of the five seconds, it is no longer moving, so yet again there is no K value. However, it is now hanging in the air, so there is Ug. Again, a spring is not involved, so there is no Us.*Note: The number of blocks representing the Ug energy does not matter as long as it is greater than 0 and the same amount that is present in the flow diagram, thereby obeying the Law of Conservation of Energy.To find the exact amount of Ug energy, we use the equation gravitational potential energy equals mass times gravity times height (Ug = m * g *h).

Ug = m * g * h

Ug = 60kg * 9.8m/s^2 * 10mUg = 5880J

We plug in everything we know, and end up with 5880J. Since it is energy, it is measured in J, or Joules.

To find power, we use the equation poewr equals work divided by time (P = w / t). We know that the work is 5880J since that is the amount of energy that was transferred during the motion, which is the definition of work. We also know time since it was given in the problem (5 seconds). Therefore, we just plug that in.

P = w / t

P = 5880J / 5s

P = 1176W

We get 1176 and use the unit watts since that is what power is measured in.