Answer:
Approximately [tex](-20.7)\; {\rm J}[/tex].
Approximately [tex]6.90\; {\rm N}[/tex].
(Assumptions: mass of the sphere is [tex]750\; {\rm g}[/tex]; gravitational field strength is [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)
Explanation:
As the sphere descends, its gravitational potential energy ([tex]\text{GPE}[/tex]) is converted into kinetic energy ([tex]\text{KE}[/tex].) If the liquid did not do any work on the sphere, the increase in [tex]\text{KE}\![/tex] should be equal to the reduction in [tex]\text{GPE}\![/tex].
However, because the liquid is resisting the motion of the sphere, in reality the increase in [tex]\text{KE}\!\![/tex] is lower than the reduction in [tex]\text{GPE}\!\![/tex]. The difference between the two is the work that the liquid did on the sphere. Note that because the force from the liquid acts opposite to the direction of motion, the work that the liquid did on the sphere would be negative.
Hence, to find the work that the liquid did on the sphere, subtract the absolute change in [tex]\text{KE}[/tex] from the absolute change in [tex]\text{GPE}[/tex].
Before calculating [tex]\text{KE}[/tex] and [tex]\text{GPE}[/tex], ensure that all quantities are in standard units:
- Mass of the sphere: [tex]m = 750\; {\rm g} = 0.750\; {\rm g}[/tex].
Assuming that the sphere was stationary the instant is was dropped. The kinetic energy of the sphere would initially be [tex]0\; {\rm J}[/tex]. At [tex]1.9\; {\rm m\cdot s^{-1}}[/tex], the kinetic energy of the sphere would be:
[tex]\begin{aligned} (\text{KE}) &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2}\, (0.750\; {\rm kg})\, (1.9\; {\rm m\cdot s^{-1}})^{2} \\ &= 1.35375\; {\rm J}\end{aligned}[/tex].
In other words, the kinetic energy of the sphere had increased by [tex]1.35375\; {\rm J}[/tex].
With a height change of [tex](-3)\; {\rm m}[/tex] (negative since the sphere would be below its initial position,) the change in the gravitational potential energy of the sphere would be:
[tex]\begin{aligned} & (\text{GPE change}) \\ =\; & m\, g\, \Delta h \\ =\; & (0.750\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}})\, (-3\; {\rm m}) \\ =\; & (-22.0725)\; {\rm J}\end{aligned}[/tex].
The absolute change in the gravitational potential energy of the sphere would be [tex]22.0725\; {\rm J}[/tex].
The difference between the reduction in [tex]\text{GPE}[/tex] and increase in [tex]\text{KE}[/tex] would be equal to the work that the liquid did on the sphere:
[tex]22.0725\; {\rm J} - 1.35375\; {\rm J} = 20.71875\; {\rm J}[/tex].
In other words, the liquid had done approximately [tex](-20.7)\; {\rm J}[/tex] of work on the sphere (negative because the liquid opposes the motion of the sphere.)
Divide the work done by the displacement in the direction of the force [tex](-3)\; {\rm m}[/tex] (negative since the force points opposite to the direction of motion) to find the average force on the sphere:
[tex]\begin{aligned}(\text{average force}) &= \frac{(\text{work done})}{(\text{displacement})} \\ &= \frac{(-20.71875\; {\rm J})}{(-3\; {\rm m})} \\ &\approx 6.90625\; {\rm N}\end{aligned}[/tex].
In other words, the average force that the liquid exerted on the sphere would be approximately [tex]6.90\; {\rm N}[/tex].
