Question

Calculate the work done in lifting a 500-N barbell 2.3 m above the floor. (what is the gain of potential energy of the barbell when it is lifted to this height?) Use g=10 m/s/s


for this question can u explain to me well why the PE is positive
i taught it would be negative

Answer 1

We will need the standard definition of work and the Work-Energy Theorem definition of work to solve this problem.

First, to calculate the work done by lifting\[W = F \cdot d = 200 \cdot 2.3\]
Second, the calculate the gain in potential energy\[W = \Delta KE + \Delta PE = 0 + mgh = 500 \cdot 2.3\]

Therefore, we can see that work equals the change in potential energy because the object is at rest after the lifting process.

To answer your second question. Anytime we raise an object above the surface of the earth, we increase it's potential energy. Take the mathematical definition of potential energy\[PE = mgh\]If h is positive, which it is anytime an object is above the surface of the earth, we have positive potential energy relative to the surface of the earth.

Answer 2

BUT g is negative

Answer 3

so we will be having -500(2)(3)

Answer 4

my prob then is that negative.since its gain in PE does it mean the negative goes away or stays

Answer 5

We take g as being positive here.

We use a negative value of g when we calculate the force of gravity and our positive y-direction is directed upwards.

Answer 6

ok thanks since we consider g positive in the upward direction.i instead took it to be negative