A 16kg sled starts up a 28 degree incline with a speed of 2.4m/s . The coefficient of kinetic friction is = 0.27.

part a)How far up the incline does the sled travel?

part b)What condition must you put on the coefficient of static friction if the sled is not to get stuck at the point determined in part (a)?

please if you can show all work thank you

(a) Initial kinetic energy =

(Potential energy increase) + (work against friction)
Let X be the distance traveled up the incline.
(M/2)V^2 = M*g*X sin28 + M*g*cos28*(0.27)X

M cancels out. Solve for X

X = (V^2/2g)/[sin28 + 0.27cos28]

(b) To make sure it slides back down the incline,the downward weight component must exceed the maximum possible static friction force.
Us*cos28*M*g < M*g*sin28
Us < tan28
(Us is the static friction coefficient)

i got 65 for part a, but it says its not right.

KE =m•v²/2,

PE =m•g•h = m•g •s•sinα,
W(fr) = F(fr) •s = k•m•g•cosα •s

KE = PE +W(fr)
v²/2 = g •s(sinα + k•cosα),
s = v²/2•g •(sinα + k•cosα)=
=0.415 m.

To solve this problem, we will use the concept of work done and energy conservation. Here's how you can find the answers:

Part a) How far up the incline does the sled travel?

1. Identify the given values:
- Mass of the sled (m) = 16 kg
- Incline angle (θ) = 28 degrees
- Initial speed (v) = 2.4 m/s
- Coefficient of kinetic friction (μk) = 0.27

2. Calculate the force of gravity (Fg) on the sled:
- Fg = m * g, where g is the acceleration due to gravity (9.8 m/s^2)
- Fg = 16 kg * 9.8 m/s^2 = 156.8 N

3. Resolve the weight of the sled into components along and perpendicular to the incline:
- F_parallel = Fg * sin(θ)
- F_perpendicular = Fg * cos(θ)

4. Determine the net external force acting on the sled:
- F_net = F_parallel - F_kinetic_friction
- F_kinetic_friction = μk * F_perpendicular

5. Calculate the work done by the net external force:
- Work = force * distance * cos(θ)
- Distance = Work / (F_net * cos(θ))

6. Substitute the known values into the equation and calculate the distance:
- Work = F_net * Distance * cos(θ)
- Distance = Work / (F_net * cos(θ))
- Distance = (F_net * v^2) / (2 * F_net * cos^2(θ))

7. Simplify the equation:
- Distance = v^2 / (2 * g * cos(θ)) - (μk * v^2) / (2 * g * cos^2(θ))

8. Substitute the given values and calculate the distance travelled:
- Distance = (2.4^2) / (2 * 9.8 * cos(28)) - (0.27 * 2.4^2) / (2 * 9.8 * cos^2(28))

Part b) What condition must you put on the coefficient of static friction if the sled is not to get stuck at the point determined in part (a)?

To ensure the sled does not get stuck at the point calculated in part a, the coefficient of static friction (μs) should be greater than the ratio of the force of gravity parallel to the incline (F_parallel) to the force of gravity perpendicular to the incline (F_perpendicular).

- μs > F_parallel / F_perpendicular
- μs > Fg * sin(θ) / Fg * cos(θ)
- μs > sin(θ) / cos(θ)
- μs > tan(θ)

Therefore, the condition for the coefficient of static friction is μs > tan(θ), where θ is the incline angle.