Question regarding potential energy?

The work #Delta W# applied to the stone is

#(f-mg)Delta h =Delta W#

this work-energy is equivalent to an additional potential energy given by

#Delta W = mg Delta H# so

#Delta H = ((f-mg)Delta h)/(m g) = ((410-32)2)/32 = 23.625# [m]

Yes, this is a question of potential energy. While the stone is in our hands, the net force on the stone does an amount of work equal to the change in the stone's kinetic energy. Once released, this kinetic energy is converted into potential energy as the stone rises.

When the kinetic energy goes to zero, the stone has reached its greatest height.

The great thing about using conservation of energy to solve this problem is that we do not actually have to calculate the kinetic energy involved. We simply calculate the work done by the net force, and convert this directly to potential energy of the stone as it rises!

Here is how:

The weight of the stone is 32 N (downward, of course), and the applied force is 410 N (upward). So, the net force on the stone is 378 N. We apply this force over a distance of 2.0 m, so the work done is

#W=F_"net"*d=(378 N) * (2.0 m) = 756 J#

The change in potential energy of the stone as it rises to a height #h# is

#DeltaU = mgh#

where #Delta U# must be equal to the work we did in throwing the stone, 756 J.

(Note that in setting the problem up this way, we are calculating the change in potential energy from the point of release to the maximum height, as required in the problem.)

Since the weight of the stone is #mg# and is equal to 32 N,

#756 = 32xxh#

So, #h = 756/ (32) = 23.625 m#

#Deltavecy' = "23.63 m"#

The idea is that the applied force over the given distance is the work performed by the surroundings (us) onto the system (the stone).

(One-dimensional) work is generally defined, for physics (in the vertical direction), as:

#w = vecF_(app)Deltavecy#

where:

• #vecF_(app)# is the force applied to the stone in #"N"#.
• #Deltavecy# is the displacement the rock went through during the performed work (including the sign), up until the rock is allowed to freely rise.

Define up as positive so that the gravitational constant #g < 0#. Then, #vecF_(app) > 0#, #vecF_g < 0# and #Deltavecy > 0#, meaning that the work performed on the stone is positive with respect to the stone, imparting kinetic energy into the stone.

This kinetic energy decreases as the stone rises into the air and its potential energy increases. Thus, when the kinetic energy reaches zero, the potential energy is at its maximum, the stone changes direction, and it begins to fall.

We are to find this maximum height the stone can travel when provided the kinetic energy imparted by the work performed and allowed to freely rise further.

Let's split this into two parts.

THE FORCED RISE

The stone has a downward force due to gravity, #vecF_g#, of #-"32 N"# as it rises, which counteracts the applied force #vecF_(app)# over the same distance #Deltavecy#, thus performing work against the surroundings.

So, the net work performed upon the stone is actually:

#w_(n et) = w_"performed" - w_"stone"#

#= "410 N"cdot"2.0 m" - "32 N"cdot"2.0 m"#

#= "756 N"cdot"m"# (or #"J"#)

This is now the kinetic energy the stone has available. That is, #K = "756 N"cdot"m"#.

THE FREE RISE

By conservation of energy, we have:

#DeltaE = 0 = K + U#,

where we set #U < 0# to account for the opposite sign of #U# (with respect to #K#).

Therefore, the magnitudes of #K# and #U# are the same, and we have:

#U = mgDeltavecy'#

#= -"756 N"cdot"m"#

where #Deltavecy'# is the further distance the stone rises.

We emphasize that #Deltavecy# and #Deltavecy'# are both positive. To obtain the further height traveled, solve for #Deltavecy'#.

#color(blue)(Deltavecy') = U/(mg)#

#= (-"756 N"cdot"m")/(-"32 N")#

#=# #color(blue)("23.63 m")#

(To not confuse the signs, which is a common confusion, recognize that there are two conventions: the sign of the energy, and the sign for a given cartesian direction.)