A child slides down a slide with a 28° incline, and at the bottom her speed is precisely three-fifths what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child

Height= length/sin28
energy available= mgh=mg length/sin28

friction loss= 2/5 of energy available
mgcos28*mu*length= 2/5 mg length/sin28

solve for mu.

Check my thinking.
Remember Sin(A+B)= 2 sinAcosB

76

Your approach to solving the problem is correct. Here's a step-by-step breakdown:

1. Start by finding the vertical height of the slide using the given angle of inclination (28°):
Height = Length / sin(28°)

2. Calculate the energy available at the top of the slide using the mass (m), acceleration due to gravity (g), and the height:
Energy available = mgh = mg * Height

3. Determine the friction loss, which is stated as three-fifths of the energy available:
Friction loss = (2/5) * Energy available

4. Apply Newton's second law in the vertical direction to find the component of the child's weight that opposes the motion:
mg * cos(28°) * μ * Length = Friction loss

5. Rearrange the equation and solve for the coefficient of kinetic friction (μ):
μ = (2/5) * (1 / (mg * cos(28°))) * Energy available / Length

Your final equation looks correct, and by solving for μ, you will get the coefficient of kinetic friction between the slide and the child.

Your thinking is mostly correct, but there are a few minor errors in your calculation. Here's the corrected version:

First, let's define the variables:
- θ = angle of inclination (28°)
- v = final speed of the child at the bottom of the slide
- v_0 = speed of the child if the slide were frictionless
- μ = coefficient of kinetic friction between the slide and the child's body
- m = mass of the child
- g = acceleration due to gravity (approximately 9.8 m/s^2)

Now, let's analyze the forces acting on the child:
1. Weight force (mg) acting vertically downward.
2. Normal force (N) acting perpendicular to the slide.
3. Frictional force (f) acting opposite to the direction of motion, parallel to the slide.

Using trigonometry, we can find the components of the weight force:
- The component parallel to the slide is mg * sin(θ).
- The component perpendicular to the slide is mg * cos(θ).

Since the child is moving with a constant velocity, the net force acting on the child must be zero:

Net force = f - mg * sin(θ) = 0

So, we can express the frictional force as:
f = mg * sin(θ)

Now, let's consider the energy lost due to friction:
The initial potential energy at the top of the slide is mgh.
The final kinetic energy at the bottom of the slide is (1/2)mv^2.

Since the child's final speed is three-fifths the speed in the frictionless case, we can write:
v = (3/5) * v_0

Now, equating the energy losses due to friction to the difference in kinetic energy, we have:
f * length_slide = (1/2)mv_0^2 - (1/2)mv^2

Plugging in the expressions for f and v, we get:
mg * sin(θ) * length_slide = (1/2)m * v_0^2 - (1/2)m * ((3/5) * v_0)^2

Simplifying the equation, we have:
μ * mg * cos(θ) * length_slide = (1/5) * m * v_0^2

Canceling out the mass terms, we find:
μ * g * cos(θ) * length_slide = (1/5) * v_0^2

Dividing both sides of the equation by g * cos(θ) * length_slide, we get:
μ = (1/5) * v_0^2 / (g * cos(θ) * length_slide)

Using the given information and trigonometric identities, we can substitute the values of θ and v_0:
μ = (1/5) * (5v^2 / 3) / (g * cos(28°) * length_slide)

Simplifying further, we have:
μ = v^2 / (3 * g * cos(28°) * length_slide)

So, to calculate the coefficient of kinetic friction (μ), you need the value of the child's final speed (v) and the length of the slide. Plug the appropriate numerals into the equation above to obtain the value of μ.