Answer:
(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
