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, [tex]\mu=6\times 10^{-2}[/tex]
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.
[tex]KE=PE+W[/tex]
[tex]\dfrac{1}{2}mv^2=mgh+W[/tex]
W is the work done by the friction.
[tex]W=f\times d[/tex]
[tex]f=\mu mg\ cos\theta[/tex]
[tex]W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}[/tex]
[tex]W=\dfrac{\mu mgh}{tan\theta}[/tex]
[tex]\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}[/tex]
[tex]\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}[/tex]
[tex]\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}[/tex]
v = 8.19 m/s
So, the speed of the box is 8.19 m/s. Hence, this is the required solution.