Question

A vehicle of weight 50 kN is moving with a speed of 60 kmph on a descending gradient of 3%. The driver of the vehicle sees an obstruction and applies the brakes. If the efficiency of the brakes is 80% and the vehicle stops at a distance of 40 m. then the frictional force developed is

• A. $17.5kN$
• B. $19kN$
• C. $21.3kN$
• D. $15kN$

Answer 1

The correct option is B $19kN$

Braking distance
$=\frac{V^2}{254(r-0.01n)}=40m$
$=\frac{{60}^2}{254(f-0.01\times 3)}=40m$
$\Rightarrow f=0.38$

Frictional force developed
$=fW=0.38\times 50=19kN$

(OR)

Breaking distance
$=\frac{V^2}{254(f\eta -G)}$
$=\frac{{60}^2}{254(f\times 0.8-0.03)}=40m$
$\Rightarrow f=0.48$

Frictional force developed
$=\eta fW=0.8\times 0.48\times 50=19kN$