Answer to Question #198923 in Mechanical Engineering for Sello

Question

The speed of a vehicle is reduced with a constant acceleration from 72km/h

72km/h to 18 km/h

km/h over 250m

250m directly down an incline of 1in25

1in25. The mass of the vehicle is 1.9 Mg

Mg and it has a constant resistance to motion of 350N

350N. 

Calculate the magnitude of the braking Force

Answer 1

Given as:

V₁=72 km/hrs=20m/s

V₂=18 km/hrs=5m/s

Distance(d)=250 m

Slope of the road ($\theta$)=1/25

Mass(m)=1900kg

Resistance force(F)=350 N

Solution:

Let $F_b$ be the braking force

Resolving weight (m×g) along the inclined road as $m\times g\times sin\theta$

Considering dynamic equilibrium along the inclined road,

$m\times g\times sin\theta -F-F_b=m\times(a)$

By kinematical equation

$(V_2)^2=(V_1)^2+2×a×d$

$a=0.75 \:m/s^2$

Substitute,

$1900\times 9.8\times sin(1/25) -350-F_b=1900\times(0.75)$

Breaking force(Fb)=1030.40 N