Physics, 20.12.2019 17:31 LILG1224

Acurve of radius 28 m is banked so that a 990 kg car traveling at 41.1 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. you are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. neglect the effects of air drag and rolling friction. if the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them?

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Answer from: theday2278

For no friction condition there is no range of speed only on possible speed is 41.1 km/h

while for the case of 0.300 friction coefficient the range of speed is from 6.5 m/s to 15.75 m/s

Explanation:

When there is no friction on the turn of road then the centripetal force is due to the component of Normal force only

So we will have

$Ncos\theta = mg$

$Nsin\theta = \frac{mv^2}{R}$

so we have

$tan\theta = \frac{v^2}{Rg}$

$\theta = tan^{-1}\frac{v^2}{Rg}$

v = 41.1 km/h = 11.42 m/s

$\theta = tan^{-1}\frac{11.42^2}{28(9.8)}$

$\theta = 25.42^o$

so here in this case there is no possibility of range of speed and only one safe speed is possible to take turn

Now in next case if the coefficient of static friction is 0.300

then in this case we have

$v_{max} = \sqrt{(\frac{\mu + tan\theta}{1 - \mu tan\theta})Rg}$

$v_{max} = \sqrt{(\frac{0.3 + 0.475}{1 - (0.3)(0.475)})(28 \times 9.8)}$

$v_{max} = 15.75 m/s$

Similarly for minimum speed we have

$v_{min} = \sqrt{(\frac{\mu - tan\theta}{1 + \mu tan\theta})Rg}$

$v_{min} = \sqrt{(\frac{-0.3 + 0.475}{1 + (0.3)(0.475)})(28 \times 9.8)}$

$v_{min} = 6.5 m/s$

So the range of the speed is from 6.5 m/s to 15.75 m/s

Answer from: Quest

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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Acurve of radius 28 m is banked so that a 990 kg car traveling at 41.1 km/h can round it even if the...

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