Engineering Mechanics

(^666) „„„„„ A Textbook of Engineering Mechanics
Now let a 1 = Retardation of the vehicle, and
t 1 = Time taken by the vehicle in coming to stop.
We know that equation for the retardation of the vehicle is
2 as=v^2 – u^2
2 a × 7·66 = 0 – (10)^2 = – 100
∴ a 1 =

• 100 –6.52m/s 2
  (2 7·66)

=

×

...(Minus sign due to retardation)

and final velocity of the vehicle (v),

0=u + a 1 t 1 = 10 – 6·52 t 1

∴ t 1 =

10

1.53 s

6.52

= Ans.

(ii) Distance covered and time taken by the vehicle in coming to stop when it is coming down the

plane

Fig. 32.9.

Let s 2 = Distance covered by the vehicle in coming to stop.

We know that as the vehicle is coming down the plane, therefore, gravitational pull due to

inclination

=Mg sin α = 2250 × 9·8 sin 10°

= 220 050 × 0·1736 = 3828 N

∴ Net resistance = 10 855 – 3828 = 7027 N

This net resistance stops the vehicle in a distance of s 2 metres. Therefore work done by the

force

= Force × Distance = 7027 s 2 ...(ii)

We know that the kinetic energy of 112 500 N-m [as per equation (i)] of the vehicle is absorbed

by the net resistance in travelling through a distance s 2. Therefore equating equation (i) and (iii),

112 500 = 7027 s 2

∴ s 2 =

112 500

16.0 m

7027

= Ans.

Now let a 2 = Retardation of the vehicle, and

t 2 = Time taken by the vehicle in coming to stop.

We know that equation for the retardation of the vehicle is

2 as=v^2 – u^2

2 a × 16 = 0 – (10)^2 = – 100