- #1
Echo88
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The problem is: A leaky 10-kg bucket is lifted from the ground to a height of
12 m at a constant speed with a rope that weighs 0.8 kg/m. Initially the
bucket contains 36 kg of water, but the water leaks at a constant rate and
finishes draining just as the bucket reaches the 12 m level. How much work
is done?
Work = Force*Distance
Force = Mass*Acceleration
Mass = Volume*Density
So, Force = Volume*Density*Acceleration
don't know how relevant some of those are...
For work that does not have a constant force
Work = the definite intregral of \int f(x)dx,
f(x) being the force.
Acceleration = 9.8 m/s^2
Density of Water = 1000 kg/m^3
I put the bounds of my integral as being [0,12]. x is the depth of the bucket
so x = 0 is the top and x = 12 is the bottom
36/12 = 3 kg/m water lost
Force = ((36-3x) + 10 + 0.8x)*9.8
Take the Integral of Force from 0 to 12
Answer = 3857.28 J
Does this look right?
12 m at a constant speed with a rope that weighs 0.8 kg/m. Initially the
bucket contains 36 kg of water, but the water leaks at a constant rate and
finishes draining just as the bucket reaches the 12 m level. How much work
is done?
Work = Force*Distance
Force = Mass*Acceleration
Mass = Volume*Density
So, Force = Volume*Density*Acceleration
don't know how relevant some of those are...
For work that does not have a constant force
Work = the definite intregral of \int f(x)dx,
f(x) being the force.
Acceleration = 9.8 m/s^2
Density of Water = 1000 kg/m^3
I put the bounds of my integral as being [0,12]. x is the depth of the bucket
so x = 0 is the top and x = 12 is the bottom
36/12 = 3 kg/m water lost
Force = ((36-3x) + 10 + 0.8x)*9.8
Take the Integral of Force from 0 to 12
Answer = 3857.28 J
Does this look right?