A steel ball of mass m falls from a height h
onto a scale calibrated in Newtons. The ball
rebounds repeatedly to nearly the same height
h. The scale is sluggish in its response to the
intermittent hits and displays an average force
Favg, such that
FavgT = F t,
where F t is the brief impulse that the ball
imparts to the scale on every hit, and T is
the time between hits. Calculate this average
force in terms of m, h, and other physical
constants.
1. Favg =mg/T
2. Favg =mg/h
3. Favg = mg
4. Favg = mgT
5. Favg = mgh

The average force F_avg = m*g

Explanation:

Given:

- Mass of the ball = m

- Initially at height = h

- Dropped from rest v_i = 0

Find:

The average force F_avg that the ball imparts to the scale on every hit.

Solution:

- The velocity of the ball just before it hits the scale can be evaluated by the energy conservation as follows:

                                  E.K_1 + E.P_1 = E.K_2 + E.P_2

- Where,                     E.K = 0 initial velocity V_i = 0

                                  E.P = 0 Scale is the datum reference.

Hence,

                                  E.P_1 = E.K_2

                                  m*g*h = 0.5*m*V_o^2

                                  V_o = sqrt ( 2*g*h)

- We will apply the conservation of momentum on the ball just before it hits the scale and just after produces an impulse. The impulse is given by Newton's second law of motion:

                                  Ft = m*(V_f - V_o )

Where, Ft is the impulse. V_f velocity after impact. V_o = V_f (no loss of energy upon impact) - assumption.

- Taking upward direction as +.

                                  Ft = m*(V_o + V_o ) = 2*m*V_o

                                  Ft = 2*m*sqrt ( 2*g*h)

- The total time between two successive hits T can be calculated by using second kinematic equation of motion in vertical direction, as follows:

                                  y = y(0) + V_o*T - 0.5*g*T^2

Where,                       y = y(0) = 0   ... start and end points same.. Datum

Hence,

                                  0 = 0+ V_o*T - 0.5*g*T^2

                                  0 = V_o - 0.5*g*T

                                  T = 2*V_o / g = 2*sqrt ( 2*g*h) / g

- The average force F_avg can now be calculated using the expression given:

                                  F_avg*T = Ft

                                  F_avg = Ft / T

                                  F_avg = [2*m*sqrt ( 2*g*h)] / [2*sqrt ( 2*g*h) / g]

                                  F_avg = m*g

bacteria that live on the ocean floor are sustained by phytoplankton. phytoplankton, microscopic marine photosynthetic organisms, have a vastly significant role to play not only in the marine food web of which they’re part of, but also on a more global scale. despite their infinitely small size in comparison to other marine organisms, these tiny creatures occupy an immensely important ecological niche: they are the foundation of the marine food web, and as primary producers, play key roles in supporting all other organisms in the marine environment, as well as in the regulation of the earth’s climate through the sequestration of carbon, oxygen production, and other related processes. phytoplankton account for roughly half of all global primary productivity; therefore, their significance extends far beyond the marine environment alone. there is an intriguing sense of irony in the realization that these tiny living beings that often live out their existence unnoticed and undetected by the rest of the world, have such a far-reaching impact on the lives of virtually all other living organisms on the planet, particularly on those in the marine environment. the world that we have become accustomed to has been and is continuously shaped by the workings of these miniscule yet vital “plants of the sea”.

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