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TiliK225 [7]
3 years ago
7

A 2020 kg car traveling at 14.2 m/s collides with a 2940 kg car that is initally at rest at a stoplight. The cars stick together

and move 2.12 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.
Physics
1 answer:
velikii [3]3 years ago
4 0

Answer:

The answer is 3,064xe^{-4}

Explanation:

When the collision happens, the momentum of the first car is applied to the both of them.

So we can calculate the force that acts on both cars as:

  • The momentum of the first car is P = 2020 kg x 14.2 m/s = 28,684 kg.m/s
  • The acceleration of both cars after the crash is going to be a = P / mtotal which will give us a = 28,684 / (2020+2940) = 5.78 m/s
  • Since the second car was initially not moving, the final acceleration was calculated with the momentum of the first car.

Now we can find the force that acts on both of them by using the formula F = m.a which will give us the result as:

  • F = (2020+2940) x 5.78 = 28,684

The friction force acts in the opposite direction and if they stop after moving 2.12 meters;

  • Friction force is Ff = μ x N where μ is the friction coefficient and the N is the normal force which is (2020+2940) x 10 if we take gravitational force as 10, equals to 49,600.
  • F - Ffriction = m x V
  • 28,684 - μ x 49,600 = 4960 x 5.78
  • μ = 3,064xe^{-4}

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Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
Fofino [41]

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

(c) vf₁ = 15.097 m/s : Stan's final speed

    vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement   

t₁ :  Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁:  Stan acceleration

d₂: car displacement   

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂:  Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ =  1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1 = equation 2

1.55*t₁²  =  2.495* t₂²  , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² = 2.495* t₂²

1.55* (t₂² +2t₂+1) = 2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0  Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant  Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁    t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂  

vf₂ =0+4.99* 3.87

vf₂ = 19.31 m/s : Kathy's final speed

8 0
3 years ago
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