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Nastasia [14]
3 years ago
10

A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.44 radians with the

horizontal. The frictional coefficients between the crate and plane are (\muμs = 0.61, \muμk = 0.23 ). What is the magnitude of the acceleration (in meters/second2) of this crate as it slides up the incline?
Physics
1 answer:
In-s [12.5K]3 years ago
6 0

Answer:

The  acceleration is  a = 6.2 m/s^2

Explanation:

From the question we are told that

    The  angle which the inclined plane make with horizontal is  \theta  =  0.44 \ rad

     The frictional coefficients are  \mu_{\mu s}  = 0.61 and  \mu_{\mu k}  =  0.23

     

The force acting on the  crate  is mathematically represented as  

        f = F_w  + F_N

Here f is the net force at which the crate is sliding down the plane which is mathematically represented as

      f = ma

        F_w is the force due to weight which is mathematically represented as

        F_w  = mg sin (\theta)

       and  F_N the force due to friction which is mathematically represented as

       F_N  =  \mu_{\mu k } * mg cos(\theta )

So  

     ma  =  mgsin(\theta ) + \mu_{\mu k} mg cos(\theta )

      a = gsin(\theta ) +  \mu_{\mu k }  *  g cos(\theta)

substituting values

      a = 9.8 sin(0.44 ) + 0.23  *  9.8* cos(0.44)

      a = 6.2 m/s^2

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Answer:

<u></u>

  • <u>2.26 seconds</u>
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<u></u>

Explanation:

1. Fall time

<u>i) Find the vertical speed of the plane when the tanks were released</u>

         V_{y,0}=84m/s\times sin(30\º)=42m/s

That is the same initial vertical speed of the tanks.

<u />

<u>ii) Find the fall time</u>

         y-y_0=V_{y,0}\cdot t+g\cdot t^2/2

         120=42t+4.9t^2

         

         4.9t^2+42t-120=0

         

        t=\dfrac{-42\pm\sqrt{(42)^2-4(4.9)(-120)}}{2\times 4.9}

Only the positive value has aphysical meaning: t = 2.26 seconds.

2. Speed when they hit the ground

<u>i) The horizontal speed is constant:</u>

          V_x=84m/s\times cos(30\º)\approx72.5m/s

<u />

<u>ii) The vertical speed is:</u>

            V_y=V_{y,0}+g\cdot t

            V_y=42m/s+9.8m/s^2\cdot (2.26s)\approx64.1m/s

<u>iii) Total speed</u>

          V=\sqrt{V_x^2+V_y^2}

          V=\sqrt{(72.5m/s)^2+(64.1m/s)^2}\approx97m/s

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Wavelength = 0.25 m

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When force is divided by area we get pressure

p=\frac{F}{A}\\\Rightarrow F=p\times A\\\Rightarrow F=7.16\times 10^4\times \pi 4\\\Rightarrow F=899752.13598\ N

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