A 62.0 kg skier is moving at 6.50 m/s on a frictionless, horizontal snow covered plateau when she encounters a rough patch 3.50 m long. The coefficient of kinetic friction between this patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.

(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?

You seem to have omitted some words of the question, along with the coefficient of kinetic friction.

Subtract the work done crossing the rough patch from the initial kinetic energy, and then add the additional kinetic energy gained when going down the frictionless hill.

From the final kinetic energy, compute the final velocity.

Whoops! Here is the complete question.

A 62.0 kg skier is moving at 6.50 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.

(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?

The work done (creating heat "internal energy") is

(Friction Force) *( distance)
= M g *0.300* 3.50 .
M is the mass and g is the acceleration of gravity.
The answer will be in Joules

Why did the skier bring a rough patch to the snow-covered plateau? To make things interesting, I guess! Now let's calculate the answers to your questions:

(a) To find the skier's speed at the bottom of the hill, we can use the conservation of energy. The potential energy she has at the top of the hill is converted to kinetic energy at the bottom. So, we set the initial potential energy equal to the final kinetic energy:

mgh = 0.5mv^2

where m is the mass of the skier (62.0 kg), g is the acceleration due to gravity (9.8 m/s^2), h is the height of the hill (2.50 m), and v is the final speed we're trying to find.

Plugging in the values, we get:

62.0 kg * 9.8 m/s^2 * 2.50 m = 0.5 * 62.0 kg * v^2

which simplifies to:

v^2 = (2 * 9.8 m/s^2 * 2.50 m)
v^2 = 49 m^2/s^2
v = √(49 m^2/s^2)
v ≈ 7.00 m/s

So the skier is moving at approximately 7.00 m/s when she reaches the bottom of the hill.

(b) Now let's calculate the Internal Energy generated in crossing the rough patch. Since the rough patch introduces friction, we can use the work-energy principle. The work done by friction will be equal to the change in kinetic energy:

Work_done_by_friction = ΔKE

The work done by friction is given by:

Work_done_by_friction = friction_force * distance

The friction force can be found using the formula:

friction_force = μ * normal_force

where μ is the coefficient of kinetic friction and normal_force is equal to the weight of the skier (mg).

The distance is given as 3.50 m.

Now, let's plug in the values and calculate:

friction_force = μ * normal_force
= μ * mg

Work_done_by_friction = friction_force * distance
= μ * mg * distance

Given the coefficient of kinetic friction (μ) and the mass of the skier (m), you can calculate the Work_done_by_friction.

Unfortunately, I don't have the specific values for the coefficient of kinetic friction and the height of the hill, which are needed to calculate the internal energy. Could you please provide those?

To solve this problem, we can use the principles of energy and motion. Let's start by breaking down the given information:

Mass of the skier (m) = 62.0 kg
Initial speed of the skier (v) = 6.50 m/s
Length of the rough patch (d) = 3.50 m
Height of the icy hill (h) = 2.50 m

(a) To find the skier's speed at the bottom of the hill, we need to consider the Law of Conservation of Energy. According to this law, the total mechanical energy of the skier remains constant if no external forces act upon her.

Initially, the skier has only kinetic energy:
Initial kinetic energy (Ek1) = (1/2) * m * v^2

When the skier reaches the rough patch, some energy is lost due to the friction. So, the skier would lose kinetic energy equal to the work done against friction (Wfriction) during the crossing of the rough patch.

Finally, when the skier reaches the bottom of the hill, all the lost energy is regained in terms of kinetic energy due to the height difference. The potential energy at the top of the hill (Ep1) is converted to kinetic energy at the bottom of the hill (Ek2).

Let's calculate each component step-by-step:

Initial kinetic energy (Ek1) = (1/2) * m * v^2
= (1/2) * 62.0 kg * (6.50 m/s)^2

Next, we need to calculate the work done against friction (Wfriction). We know that work done against friction is equal to the force of friction multiplied by the distance traveled. The force of friction is given by the equation:

Force of friction (Ffriction) = coefficient of kinetic friction (μ) * normal force (Fnormal)

We can assume that the normal force is equal to the weight of the skier, as she is on a horizontal surface. Therefore, Fnormal = m * g, where g is the acceleration due to gravity (9.8 m/s^2).

The formula for work is:
Work done against friction (Wfriction) = Ffriction * d

Substituting the above equations, we can calculate the work done against friction:

Wfriction = (μ * Fnormal) * d
= (μ * m * g) * d

Now, let's calculate the potential energy at the top of the hill (Ep1) and the total mechanical energy at the bottom of the hill:

Potential energy at the top of the hill (Ep1) = m * g * h
Total mechanical energy at the bottom of the hill = Ep1 + Ek2

Since energy is conserved, the total mechanical energy at the bottom of the hill should be equal to the initial kinetic energy (Ek1 - Wfriction):

Total mechanical energy at the bottom of the hill = Ek1 - Wfriction

Finally, we can solve for the velocity at the bottom of the hill:

Total mechanical energy at the bottom of the hill = Ek2
Ek1 - Wfriction = (1/2) * m * v^2

(b) To find the internal energy generated while crossing the rough patch, we can determine the work done against friction (Wfriction) as calculated earlier. The internal energy generated is equal to the work done against friction:

Internal energy generated = Wfriction

Now that we have broken down the problem and derived the necessary equations, we can substitute the given values into the equations and compute the answers.