Starting at rest at the edge of a swimming pool, a 72.0 kg athlete swims along the surface of the water and reaches a speed of 1.10 m/s by doing the work Wnc1 = +165 J. Find the nonconservative work, Wnc2 done by the water on the athlete.

Thanx in advance.

Well, isn't it funny how athletes can swim and work at the same time? But let's dive into the question anyway!

The work done by the athlete, Wnc1, is given as +165 J. Since it has a positive sign, we can assume that it's done by an external nonconservative force, like friction or some other external factor.

Now, to find the work done by the water on the athlete, Wnc2, we can use the work-energy theorem, which states that the work done by all forces (conservative and nonconservative) is equal to the change in kinetic energy.

Since the athlete starts at rest, the initial kinetic energy is zero. So, the equation becomes:

Wnc1 + Wnc2 = ΔKE

Now, let's substitute the values given in the problem. The mass, m, is 72.0 kg, and the final velocity, vf, is 1.10 m/s. Thus, the change in kinetic energy, ΔKE, is:

ΔKE = (1/2)m(vf^2 - vi^2)

But since the athlete starts from rest, vi = 0. Therefore:

ΔKE = (1/2)m(vf^2 - 0^2)
= (1/2)m(vf^2)

Now, plugging in the values:

165 J + Wnc2 = (1/2)(72.0 kg)(1.10 m/s)^2

Calculating that out, we find that:

165 J + Wnc2 = 43.38 J

To solve for Wnc2, we just need to isolate it:

Wnc2 = 43.38 J - 165 J

And the answer is:

Wnc2 = -121.62 J

So, the water does negative 121.62 Joules of work on the athlete. Maybe the water was feeling a bit lazy and decided to hinder the athlete's progress.

To find the nonconservative work done by the water on the athlete, you need to use the work-energy theorem. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.

The work done by the nonconservative forces is represented by the equation:

Wnc2 = ΔKE - Wnc1

where Wnc2 is the nonconservative work done by the water on the athlete, ΔKE is the change in the kinetic energy of the athlete, and Wnc1 is the nonconservative work done by the athlete.

Given that the athlete starts at rest and reaches a speed of 1.10 m/s, we can calculate the change in kinetic energy:

ΔKE = (1/2) * m * (vf^2 - vi^2)

where m is the mass of the athlete, vf is the final velocity, and vi is the initial velocity. Plugging in the values:

ΔKE = (1/2) * 72.0 kg * (1.10 m/s)^2 - 0

Simplifying the equation:

ΔKE = 0.5 * 72.0 kg * 1.21 m^2/s^2
ΔKE = 43.56 J

Now, we can substitute the values into the initial equation to calculate the nonconservative work done by the water on the athlete:

Wnc2 = ΔKE - Wnc1
Wnc2 = 43.56 J - 165 J
Wnc2 = -121.44 J

Therefore, the nonconservative work done by the water on the athlete is -121.44 J. The negative sign indicates that the work is done against the direction of motion.

To find the nonconservative work done by the water on the athlete, we can use the work-energy principle. This principle states that the work done on an object is equal to the change in its kinetic energy. In equation form:

Wnc = ΔKE

In this case, the athlete is initially at rest, so their initial kinetic energy is zero. We are given the final speed of the athlete, which is 1.10 m/s, so we can calculate the final kinetic energy using the formula:

KE = (1/2)mv^2

where m is the mass (72.0 kg) and v is the final speed (1.10 m/s). Plugging in these values, we get:

KE = (1/2)(72.0 kg)(1.10 m/s)^2
= 43.56 J

Now we can find the nonconservative work by subtracting the initial kinetic energy (zero) from the final kinetic energy:

Wnc2 = ΔKE = KE - KE_initial
= 43.56 J - 0 J
= 43.56 J

Therefore, the nonconservative work done by the water on the athlete is +43.56 J.