A bucket that has a mass of 30 kg when filled with sand needs to be lifted to the top of a 30 meter tall building. You have a rope that has a mass of 0.3 kg/m that must be secured to the bucket. It takes 1 meter of rope to secure the bucket. Once the bucket reaches the top of the building it only has mass 15 kg because there was a hole in the bottom and sand was leaking out at a constant rate while it was being lifted to the top of the building. Find the work done lifting the bucket (sand and rope) to the top of the building.

765 J

Step-by-step explanation:

We are given;

Mass of bucket = 30 kg

Mass of rope = 0.3 kg/m

height of building= 30 meter

Now,

work done lifting the bucket (sand and rope) to the building = work done in lifting the rope + work done in lifting the sand

Or W = W1 + W2

Work done in lifting the rope is given as,

W1 = Force x displacement

W1 = (30,0)∫(0.2x .dx)

Integrating, we have;

W1 = [0.2x²/2] at boundary of 30 and 0

W1 = 0.1(30²)

W1 = 90 J

work done in lifting the sand is given as;

W2 = (30,0)∫(F .dx)

F = mx + c

Where, c = 30 - 15 = 15

m = (30 - 15)/(30 - 0)

m = 15/30 = 0.5

So,

F = 0.5x + 15

Thus,

W2 = (30,0)∫(0.5x + 15 .dx)

Integrating, we have;

W2 = (0.5x²/2) + 15x at boundary of 30 and 0

So,

W2 = (0.5 × 30²)/2) + 15(30)

W2 = 225 + 450

W2 = 675 J

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,

W = 90 + 675

W = 765 J

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

look at the picture.

y-intercept for the function on a picture is 4 → point (0, 4).

y-intercept for the function

is for x = 0. substitute:

y-intercept is 8 → point (0, 8).