How is potential energy related to work?

Question

A student lifts a hammer, which has a weight of 5 N, a vertical distance of 0.8 m. Calculate
a) the work the student does on the hammer,
b) the increase in the hammer's energy.

I calculated the first part. I came up with 4 joules using the formula:
work (joules) = force (newtons) x distance (meters)
work = 5N x 0.8m
work = 4 joules

I have no idea what to do with the second question.

Answer 1

Your answer to a) correct.



Based on the simplicity of this question I`m going to assume that the student moved the hammer from rest and that it ended at rest. Now we can use conservation of energy to get b). The student did 4J of work and put 4J of his mechanical energy into the system. Since we assumed that the hammer is not in motion (no kinetic energy) and is not emitting energy in the form of sound or something, then by conservation of energy the mechanical energy of the student must have been transferred to potential energy of the hammer - so the answer to b) is 4J.



Alternatively you can use U=mgh and take the initial position of the hammer to be that of zero potential energy. Thus U = 5N*0.8m = 4J.

Answer 2

am i in a position to furnish you extremely tip. sometime, whenever you wright an equation utilising letters, be sure which you in basic terms continually state for sure what style each letter stands for. specific text fabric books in many situations use distinctive letters for the equivalent parts. So, is Q a cost in Coulombs? Is V a voltage in volts? If that's superb, then your first equation is improper. If E is the E-self-discipline then the rigidity of the area could be written in 2 techniques E- self-discipline = Volts / meter or force / Coulomb E-self-discipline = V / m or N / C The equation E = (0.5 of) m v^2 is the equation for an products kinetic power, the place E = power in joules; m = mass in Kg ; and v = speed in m/s So i think you have gotten gained some equations blended up superb right here. The power of a fee in an electric powered area is given by En = (a million/2) QV the place En = power in Joules; Q = fee in coulombs; V = voltage in volts V = 2En/Q shop in mind that the magnitude of the cost on an electron is a million.6 * 10^-19 C V = (2 * 4.5 * 10^-sixteen) / (a million.6 * 10^-19 ) V = 5.625 * 10^3 Volts

Answer 3

work and potential energy aren't necessarily related (eg you could be doing work in moving an object horizontally and have no increase in potential energy). Potential Energy (Epot) = mass * acceleration due to gravity * height, which is also equal to weight * height so its increase in Epot is 5N*.8m.