You are a member of an alpine rescue team and must get a box of supplies, with mass 2.70 kg , up an incline of constant slope angle 30.0 ∘ so that it reaches a stranded skier who is a vertical distance 2.90 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00×10−2. 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 . Part A 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.

The work-energy therom says that the total work done by external forces goes to change the kinetic energy of a system. Here you have to print to the body an inital kinetic energy that at the end will be transferred to an increase of the potential energy of the body and dissipated in friction.

So, [tex]1/2*m*(v_{f}^2-v_{i}^2)=m*g*\Delta h + Wr[/tex]

where [tex]\Delta h[/tex] is the height of the slope (2.9 m) and Wr is the frictional work. To account for the minimum energy, we have to consider that the final velocity is zero.

To account for friction, the normal force Fr=m*g*[tex]\mu[/tex]=2.7kg * 9.81m/s^2 * cos(30º) * 0.06=1.38 N.

The total length of the slope (d) is equal to 2.9/sin(30º)=5.8 m.. Thus, the work done by friction forces Wr=1.38 N * 5.8 m=8.004 J.

Finally,

[tex]1/2 * 2.7 kg *v_{i]^2=2.7 kg *9.81m/s^2 *2.9 m + 8.004 J[/tex]

[tex]v_i=\sqrt{\frac{2*84.82 J}{2.7 m}}=7.93 m/s[/tex]