In the school cafeteria, a trouble-making child blows a 12.0 g spitball through a 25.0 cm straw. The force

of her breath as a function of the distance along the length of the straw can be modeled as 10-50x^2-70x^3
where force is in newtons and x is in meters. (a) How much work is done by her breath on the spitball as it travels the length of the barrel? (b) Assuming negligible friction and the straw is held horizontally, what is the speed of the spitball as it leaves the straw?

F = 10 - 50x^2 - 70x^3.

F=10 - 50*(0.25)^2 - 70*(0.25)^3=5.78 N.

a. W = F * d = 5.78 * 0.25=1.45 Joules.

b. a = F/m = 5.78 / 0.012 = 481.7 m/s^2.
V^2 = Vo + 2a*d.
V^2 = 0 + 963.4*0.25 = 240.85
V = 15.5 m/s.

This is incorrect on part A. Work is a measurement of force over position / time, and since the force of the spitwad as it moves through the straw is a function of it's position, the force at the initial position 0 is 10 Newtons, and then ~5 - 25/32 Newtons or so at 0.25 meters. To find work, integrate the function of the force from position 0 to position 0.25.

Integral (10-50*x^2-70*x^3, x, 0, 0.25)

The area under the force curve gives you the work done for all changing points from 0 to 0.25.

As a note, multiplying the force by the change in distance is only applicable if the force is constant. Multiplying a constant linear force by the change in distance gives you the area under the constant force curve.

To solve part (a), we need to calculate the work done by the force of her breath on the spitball as it travels the length of the straw. The work done can be found using the formula:

Work = Force x Distance

In this case, the force as a function of distance along the straw is given by:

F(x) = 10 - 50x^2 - 70x^3

We are asked to find the work done as the spitball travels the entire length of the barrel, which is 25 cm or 0.25 m.

To calculate the work, we integrate the force function with respect to distance from 0 to 0.25:

Work = ∫[0 to 0.25] (10 - 50x^2 - 70x^3) dx

Now, let's solve this integral step by step:

∫ (10 - 50x^2 - 70x^3) dx
= ∫ 10 dx - ∫ 50x^2 dx - ∫ 70x^3 dx
= 10x - (50/3)x^3 - (70/4)x^4

Evaluating this expression from 0 to 0.25 gives us the total work done:

Work = [10(0.25) - (50/3)(0.25)^3 - (70/4)(0.25)^4] - [10(0) - (50/3)(0)^3 - (70/4)(0)^4]

Work = 2.0833 J

Therefore, the work done by her breath on the spitball as it travels the length of the barrel is approximately 2.0833 Joules.

Now, let's move on to part (b) to determine the speed of the spitball as it leaves the straw.

We will use the work-energy principle to find the speed of the spitball. The work done on the spitball is equal to its change in kinetic energy.

The work done (which we calculated in part (a)) is equal to:

Work = Change in Kinetic Energy

As the spitball starts from rest inside the straw, its initial kinetic energy is zero. Therefore, the work done is equal to its final kinetic energy:

Work = Kinetic Energy

The final kinetic energy of the spitball can be written as:

Kinetic Energy = (1/2)mv^2

Where m is the mass of the spitball and v is its final velocity.

Since the mass of the spitball is given as 12.0 g, which is equal to 0.012 kg, and the work done is 2.0833 J, we can set up the equation:

2.0833 J = (1/2)(0.012 kg)(v^2)

Simplifying the equation:

v^2 = (2 * 2.0833 J) / (0.012 kg)
v^2 = 347.22 m^2/s^2
v = sqrt(347.22)

Calculating the square root:

v ≈ 18.64 m/s

Therefore, the speed of the spitball as it leaves the straw is approximately 18.64 m/s.