You are a member of an alpine rescue team and must get a box of supplies, with mass 3 kg , up an incline of constant slope angle 30.0, so that it reaches a stranded skier who is a vertical distance 4 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 0.05. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s2 .What is the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier

For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.

W = Em_f - Em₀ (1)

Starting point. Lower slope

Em₀ = K = ½ m v²

highest point. Where is the skier at a height h

Em_f = U = m g h

The work of rubbing

W = -fr x

the negative sign is because the friction force opposes the movement.

Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular

let's use trigonometry to break down the weight

cos θ = W_y / W

sin θ = Wₓ / W

W_y = W cos θ

Wₓ = W sin θ

Y axis

N - Wₓ = 0

N = mg sin θ

X axis

fr = m a

the friction force has the expression

fr = μ N

fr = μ mg sin θ

we look for the job

W = - μ mg sin θ x

where x is the distance along the slope

we substitute in 1

-μ mg sin θ x = mg h - ½ m v²

let's use trigonometry to find the distance x

tan 30 = h / x

x = h / tan 30

we substitute

- [tex]\mu \ mg \ sin \theta \ \frac{h}{tan 30} \ x[/tex] = m gh - ½ m v²

we use

tan 30 = sin30 / cos30

v² = 2g h + 2 μ g h cos 30

v = [tex]\sqrt{ 2gh \ (1+ cos 30}[/tex]

let's calculate

v = [tex]\sqrt{ 2 \ 9.8 \ 4 \ (1 + 0.05 \ cos \ 30)}[/tex]

v = 9.04 m / s