Question on Work and Energy

[atom]

Please help me with this question:

How much work is done on an object of 4.5kg when it is lifted vertically at a constant speed through a displacement of 6.0m?

Thanks in advance!

[Jordan.Carroll]

Work = F . x

Since acceleration = 0
Sum of forces = 0

Therefore force down (Mg) has to equal force up.
Therefore
F=mg

x= displacement, and since it is not at an angle to the force, you don't need to include a trig function
so x=6

U=F . x             Let g=10
U=(mg)
U=(4.5*10) * 6
U= 45 * 6
U=270
work done is 270 joules

[yawho]

How much work is done on an object of 4.5kg when your supporting hand allows the object to move down at a constant speed through a displacement of 6.0m?

[Jordan.Carroll]

Same thing, since v= constant
 dv/dt=0
 so acceleration = 0.

sum of forces = 0

therefore force up= force down
Force down = mg
Force up= - mg
U=F . x
=9.81 * 4.5 * 6
 Simplistically.

However, as you lower your hand at a constant speed, there will be a certain amount of distance that the speed of your hand will be faster than the speed of the object. The dropped object will have to catch up to your hand.

The distance travelled by your hand will be modelled on

x=vt

and the distance travelled by the mass will be

x=1/2 at^2
x=.5 g t^2

when these distances are the same the mass will be touching your hand, and it will not be accelerating and your hand will be exhibiting a constant force of Mg upwards on the mass.

vt=g/2 t^2
gt^2-2vt=0
t^2-(2/g)vt=0
t(t-(2/g)v)=0

as you can see there are two solutions for this; t=0, as we expect because you are holding it before you start to move, and at t=(2/g)v

Since we know what t is, we can substitute it back into our distance equation.

vt=x
v ((2/g) v) = x
2v^2/g=x

So distance travelled against the force is the total distance minus the distance travelled where the object is in free fall
 = (6 - 2v^2/g)

Since U= F.x
U= mg ( 6- 2v^2/g)
U=6mg -2mv^2g/g
U=6mg -2mv^2
U=2m(3g-v^2)
U=9(3g-v^2)