Question

Relative to the floor, what potential energy does a 2.5 kg package have that sits on a shelf 2.2 m high? How much work was done to give it this PE?

Answer 1

Who is right?

Teachers Answers:
1. 53.9 J for both Questions

My Answers:
1. 54 J for the first one and 119J for the second question

PEg = mgh
= (2.5)(9.8)(2.2)
= 53.9 J ----> 54 J

W = F * D
= (54) (2.2)
= 119 J

Answer 2

Your teacher is right. The potential represent the energy that is required to lift the package up to 2.2 meter high. D is displacement. Lets say you lift the package from the floor up to 2.2 meter high but then bought it back down to the floor. The displacement is zero and therefore no work was done.

Answer 3

@ospreytriple are you okay typying or what lol?

Answer 4

The work-potential energy theorem applies here:
\[E_g = mg \Delta h\]\[=\left( 2.5 \right)\left( 9.81 \right)\left( 2.2 \right)\]\[=54 \text{ J}\]\[W=\Delta E_g\]\[=E_{g,f} - E_{g,i}\]\[=54 \text{ J} - 0\]\[=54 \text{ J}\]Your teacher is incorrect for the second question. The force is not 54 N as shown. The force required to lift the package is its weight, i.e.
\[F=mg\]\[=\left( 2.5\text{ kg} \right)\left( 9.81\text{ m/s^2} \right)\]\[=24.5 \text{ N}\]So another way to attack this problem is as follows (and more like your teacher's solution);
\[W=F \Delta d\]\[=mg \Delta d\]\[=\left( 2.5 \right)\left( 9.81 \right)\left( 2.2 \right)\]\[=54 \text{ J}\]Look familiar? It's exactly the result you get by applying the work-potential energy theorem.

Answer 5

which way do I pick for the second question

Answer 6

My advice is to use whatever approach your teacher has taught you. BTW, I misread your earlier post. Your teacher's answers are correct. You made the error in the second questions. The force is not 54 N, but the weight of the object.