Respuesta :
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
[tex]F_f = \mu N[/tex]
[tex]F_f = \mu m g cos \theta[/tex]
work against friction
W = F d
[tex]W = \mu m g cos \theta \times h l sin\theta[/tex]
[tex]W = \dfrac{\mu m g \times h}{tan\theta}[/tex]
Potential energy
PE = mgh
[tex]\dfrac{1}{2}mv^2= \dfrac{\mu mgh}{tan \theta}+ mgh[/tex]
[tex]\dfrac{1}{2}v^2= \dfrac{0.06 \times 9.81 \times 3.3}{tan 30^0}+ 9.8\times 3.3[/tex]
v = 8.45 m/s
the minimum speed is equal to 8.45 m/s
We have that for the Question "" it can be said that the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier is
- [tex]v=8.45m/s\\\\[/tex]
From the question we are told
- 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.
Generally the equation for total radiation acceleration is mathematically given as
[tex]a=ugcos30+gsin30\\\\a=6.0*10^{-2}(9.8)cos30+9.8sin30\\\\a=5,41m/s^2\\\\[/tex]
Generally the equation for speed is mathematically given as
[tex]v=\sqrt{2ah}\\\\v=\frac{3.3}{sin30}\\\\v=6.6m\\\\[/tex]
Therefore
[tex]v=\sqrt{2ah}\\\\v=\sqrt{2*5.41*6.6}\\\\v=8.45m/s\\\\[/tex]
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