Question

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope angle 30.0 ∘ so that it reaches a stranded skier who is a vertical distance 3.10 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 .
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:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

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

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.