Question

You are a member of an alpine rescue team and must get a box of supplies, with mass 3.00kg , up an incline of constant slope angle 30.0∘ so that it reaches a stranded skier who is a vertical distance 2.90m 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 9.81m/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.

Answer 1

L = h/sinA = 2.90/sin30 = 5.80 m. = Length of incline.

Wb = = m*g = 3kg * 9.8N/kg = 29.4 N. = Weight of box.

Fp = 29.4*sin30 = 14.7 N. = Force parallel to incline.

Fv = 29.4*cos30 = 25.46 N. = Force
perpendicular to incline.

Fk = u*Fv = 0.06*25.46 = 1.53 N. = Force of kinetic friction.

PE + KE = mg*h-Fk*L
0 + 0.5m*Vo^2 = mg*h-Fk*L
1.5Vo^2 = 29.4*2.9-1.53*5.8 = 76.39
Vo^2 = 50.9
Vo = 7.14 m/s. = Initial Velocity.