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3 years ago

15

Someone please help me answer these

1 answer:

nekit [7.7K]3 years ago

4 0

1. GPE

2. KE

3. KE

4. KE

5. Both

6. Both

7. Neither

8. Neither

Alright I think these should be right ;)

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1 year ago

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2 years ago

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

so

$\theta=5.71^{o}$

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2 years ago