Acar has a mass of 1520 kg. while traveling at 20 m⁄s, the driver applies the brakes to stop the car on a wet surface with a 0.40 coefficient of friction. (a) how far does the car travel before stopping? (b) if a different car with a mass 1.5 times greater is on the road traveling at the same speed and the coefficient of friction between the road and the tires is the same, what will its stopping distance be? explain your results.

(a)d₁ = 51.02 m

(b)d₂ =51.02m

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Known data

m=1520 kg  : mass of the  car

μk= 0.4 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the car

We define the x-axis in the direction parallel to the movement of the  car and the y-axis in the direction perpendicular to it.

W: Weight of the block : In vertical direction  downward

FN : Normal force :  In vertical direction  upward

f : Friction force:  In horizontal direction  

Calculated of the W

W= m*g

W=  1520 kg* 9.8 m/s² = 14896 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN - Wy = 0

FN = Wy

FN = 14896 N

Calculated of the f

f = μk* N= (0.4)* (14896 N )

f = 5958.4 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

- f = m*a

-5958.4 = (1520)*a

a  =  (-5958.4) /  ((1520)

a = -3.92 m/s²

(a) displacement of the car (d₁)

Because the car moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d₁ Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 20 m⁄s

vf = 0

a = --3.92 m/s²

We replace data in the formula (2)  to calculate the distance along the ramp the block reaches before stopping (d₁)

vf²=v₀²+2*a*d ₁

0 = (20)²+2*(-3.92)*d ₁

2*(3.92)*d₁  = (20)²

d₁ = (20)² / (7.84)

d₁  = 51.02 m

(b)  Different car

m₂ = 1.5 *1520 kg

μk₂= 0.4

W₂= m*g

W₂=   (1.5) *1520 kg* 9.8 m/s² = (1.5)*14896 N  

FN₂=  (1.5)*14896 N  

f= 0.4* (1.5)*14896 N  

a = - f/m₂ = - 0.4* (1.5)*14896 N  /(1.5) *1520

a = -3.92   m/s²

vf²=v₀²+2*a*d₂

vf=0 , v₀=20 m⁄s , a = -3.92   m/s²

d₂ = d₁ = 51.02m

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

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