Question

You are a member of an alpine rescue team and must get a box of supplies, with mass 3.00 kg , up an incline of constant slope angle 30.0° so that it reaches a stranded skier who is a vertical distance 3.30 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/s^2 . 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

Answer:

v = 8.45 m/s

Explanation:

given,

mass  = 3 kg

angle = 30.0°

vertical distance = 3.3 m

μ = 0.06

according to conservation of energy

KE(loss) = PE(gain) + Work done (against\ friction)..............(1)

frictional Force

$F_f = \mu N$

$F_f = \mu m g cos \theta$

work against friction

W = F d

$W = \mu m g cos \theta \times h l sin\theta$

$W = \dfrac{\mu m g \times h}{tan\theta}$

Potential energy

PE = mgh

$\dfrac{1}{2}mv^2= \dfrac{\mu mgh}{tan \theta}+ mgh$

$\dfrac{1}{2}v^2= \dfrac{0.06 \times 9.81 \times 3.3}{tan 30^0}+ 9.8\times 3.3$

v = 8.45 m/s

the minimum speed is equal to 8.45 m/s