You are a member of an alpine rescue team and must get a box of supplies, with mass 2.50 kg, up an incline of constant slope angle 30.0° so that it reaches a stranded skier who is a vertical distance 3.50 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00x102. 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/s Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. Express your answer numerically, in meters per second.
1. How to approach the problem
2. Find the total work done on the box
3. Initial kinetic energy
4. What is the final kinetic energy?

v₀ = 2.67 m / s

Explanation:

This problem can be solved using the Kinetic Enemy Work Theorem

         W = ΔK

Work is defined by the relation

         W = fr. d

The bold letters indicate vectors, in this case the blow is in the direction of the slope of the ramp and the displacement is also in the direction of the ramp, therefore the angle between the force and the displacement is zero.

the friction force opposes the displacement therefore its angle is 180º

          W = - fr d

Let's use Newton's second law, we define a reference frame with the horizontal axis parallel to the plane

Y axis  

           N- Wy = 0

           N - W cos tea = 0

the friction force has the expression

          fr = μ N

          fr = μ W cos θ

we substitute

           W = - μ W cos θ d

let's look for kinetic energy

the minimum velocity at the highest point is zero

           K_f = 0

the initial kinetic energy is

            K₀ = ½ m v₀²

we substitute energy in the work relationship

         - μ W cos θ d = 0 - ½ m v₀²

           v₀² = - μ W cos θ  2d / m

Let's use trigonometry to find distance d

         sin θ=  y / d

         d = y /sin θ

         d = 3.50 / sin 30

         d = 7 m

let's calculate

           v₀² = (6 10⁻² 2.50 9.8 cos 30) 2 7 / 2.50

           v₀ = √7.129

           v₀ = 2.67 m / s

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

the answer is b, have negligible mass