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Romashka [77]
3 years ago
14

Determine the minimum angle at which a frictionless road should be banked so that a car traveling at 20.0 m/s can safely negotia

te the curve if the radius of the curve is 200.0 m.

Physics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

Minimum angle at which frictionless road should be banked is 11.53°

Explanation:

Consider the figure below to understand the banking curve. θ is  minimum angle at which frictionless road should be banked.

Forces acting along x-axis:

F_{N}sin\theta=\frac{mv^{2}}{R}---(1)\\\\

Forces acting along y-axis:

F_{N}cos\theta-mg=0---(2)\\\\

Dividing (1) and (2)

\frac{F_{N}sin\theta}{F_{N}cos\theta}=\frac{\frac{mv^{2}}{R}}{mg}\\\\tan\,\theta=\frac{v^{2}}{rg}\\\\v=20\,m/s,\,r=200\,m\\\\tan\,\theta=\frac{(20)^{2}}{(200)(9.8)}\\\\tan\,\theta=0.204\\\\\theta=tan^{-1}(0.204)\\\theta=11.53^{o}\\

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The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of
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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is  =4.51m/s  

Explanation:

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                v^2 = \frac{2(112.86 -\mu_kmgx)}{m}

                v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}

and we are told that coefficient of friction  = 0.4 and the mass is 9 kg ,the acceleration due to gravity = 9.8m/s^2  this displacement length of spring = 0.6

  Therefore   v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }

                        =4.51m/s      

           

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