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Burka [1]
3 years ago
12

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 an

gle 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.
Physics
1 answer:
a_sh-v [17]3 years ago
8 0

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.

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4 0
2 years ago
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A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti
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Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

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Now,

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3 years ago
two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y
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Answer:

Explanation:

Given

initial velocity component of engines is

v_0_x=6380 m/s

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time period of engine running=763 s

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y=7.27\times 10^6

Using s=ut+\frac{at^2}{2} in x and y direction

x=v_0_x\times t+\frac{at^2}{2}

4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}

4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}

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In y direction

y=v_0_y\times t+\frac{a't^2}{2}

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y component=7.24 m/s^2

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3 years ago
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