In the school cafeteria, a trouble-making child blows a 12.0 g spitball through a 25.0 cm straw. The force (F) in Newtons, of his breath as a function of the distance along the straw (x) in meters, can be modeled as a linearly decreasing function for the first half of the straw, followed by a constant force through the rest of the straw. The force decreases by half along the first half of the straw. Assume there is negligible friction and the straw is held horizontally.
(a) Sketch a plot of the force of his breath as a function of position along the straw, labeling the force at x = 0 as F0.
(b) If the spitball begins from rest and leaves the straw with a speed of 16 m/s, how much work is done on the spitball?
(c) What is the maximum force F0, that acts on the spitball?

Given that the force (F) in Newtons, of his breath as a function of the distance along the straw (x) in meters, can be modeled as a linearly decreasing function for the first half of the straw,

Half of the straw = 12.5 cm

Let Force F = dependent variable

And distance x = independent variable

A.) Find the attached file for the graph.

Sinnce the force decreases by half along the first half of the straw. Assume there is negligible friction and the straw is held horizontally

B.) Given that the mass M = 12g = 12/1000 = 0.012kg

And Velocity V = 16m/s

The workdone = the kinetic energy of the split ball

WD = 1/2mv^2

Substitutes m and V into the formula

WD = 1/2 × 0.012 × 16^2

WD = 1.536 Joule

C.) The maximum force acts on the spit ball will be at maximum kinetic energy.

Since work done = force × distance

Where distance = 12.5 cm

F × S = 1/2mv^2

Substitutes all the parameters

F × 0.0125 = 1.536

Make F the subject of formula

F = 1.536/0.0125

F = 122.88 N

The maximum force F0, that acts on the spitball is therefore equal to 123 N approximately.

okpalawalter8 okpalawalter8 · Answer 2 · 2021-12-01T13:16:44+01:00

We have that for the Question "(a) Sketch a plot of the force of his breath as a function of position along the straw, labeling the force at x = 0 as F0.

(b) If the spitball begins from rest and leaves the straw with a speed of 16 m/s, how much work is done on the spitball?

(c) What is the maximum force F0, that acts on the spitball?"

It can be said that

a) The sketch is attached

b) The workdone = [tex]1.536J[/tex]

c) Maximum force = [tex]6.4N[/tex]

From the question we are told

mass of spitball = 12g through a 25.0 cm straw

Generally the equation for workdone is mathematically given as

[tex]= \frac{1}{2}mv^2 - 0\\\\= \frac{1}{2} * 12*10^{-3} * 16^2\\\\= 1.536J[/tex]

Maximum force

[tex]w = \frac{1}{2}*\frac{x_0}{2}*\frac{F_0}{2} + x_0*\frac{F_0}{2}\\\\w = \frac{5}{8}x_0F_0[/tex]

Therefore,

[tex]F_0 = \frac{8w}{5x_0}\\\\F_0 = \frac{8*1.536}{5*25*10^[-2]}\\\\F_0 = 6.4N[/tex]

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