Question

Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approaches a long embankment inclined upward at 16° above the horizontal. As she slides up the embankment, she encounters a coefficient of friction of 0.128. Determine the height to which she will travel before coming to rest.

Answer 1

To find the height Sheila will ascend before stopping, we use energy conservation, calculating her initial kinetic energy and the work done against friction to solve for the final potential energy, which includes the height she reaches on the incline.

Step-by-step explanation:

To determine the height Sheila will travel up the embankment before coming to rest, we need to apply the principles of energy conservation and include the work done against friction. Sheila's initial kinetic energy (KE) at the bottom of the hill is converted into gravitational potential energy (PE) and the work done against friction as she moves up the incline.

We can calculate the initial kinetic energy using the formula KE = 1/2 * m * v2, where m is her mass and v is her velocity. Then, we find the work done against friction, which is equal to the force of friction times the distance traveled (Wfriction = Ffriction * d). The force of friction is calculated as the coefficient of friction multiplied by the normal force, which, on an incline, is the component of the gravitational force perpendicular to the slope.

Using these concepts, we can set up the equation: KEinitial = PEfinal + Wfriction. Solving for the final potential energy gives us PEfinal = m * g * h, where h is the height she will reach. The height can be determined by rearranging this equation after calculating the work done against friction and knowing the initial kinetic energy.

It is important to note that since no values are given for distances or the length of the incline and we are assuming a constant coefficient of friction, we find the height as a function of distance she travels up the slope until she comes to rest.

Answer 2

y = 54.9 m

Step-by-step explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis

N - W cos θ = 0

The equation for the force of friction is

fr = μ N

fr = μ mg cos θ

We replace at work

W = - μ m g cos θ d

Mechanical energy in the lower part of the embankment

Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

`Em_(f)` = U = m g y

W = ΔEm =
`Em_(f)` - Em₀

- μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

sin θ = y / d

y = d sin θ

- μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

d (g syn θ + μ g cos θ) = ½ v²

d = v²/2 g (sintea + myy cos tee)

d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

d = 1555.85 /7.8145

d = 199.1 m

Let's use trigonometry to find the height

sin 16 = y / d

y = d sin 16

y = 199.1 sin 16

y = 54.9 m