Engineering Mechanics

Chapter 30 : Work, Power and Energy „„„„„ 615

Solution. Given : Mass of the wagon (m) = 50 t ; Initial velocity (u) = 0 (because it starts
from rest) ; Distance (s) = 30 m ; Gradient (sin θ) = 1% = 0.01 ; Track resistance = 50 N/t
= 50 × 50 = 2500 N = 2.5 kN and bumper spring constant (k) = 1 mm/ 20 kN = 0.05 mm/ kN

Velocity with which the wagon strikes the post

Let v = Velocity with which the wagon strikes the post.

Since the wagon is travelling down the grade, therefore gravitational pull

= mg sin θ = 50 × 9.8 × 0.01 = 4.9 kN

∴Net force responsible for moving the wagon

= Gravitational pull – Tractive resistance

= 4.9 – 2.5 = 2.4 kN

We know that net force responsible for moving the wagon

2.4 = ma = 50 a

∴

2.4 0.048 m/s 2

50

a==

We also know that relation for the velocity of the engine,

v^2 = u^2 + 2 as = (0)^2 + 2 × 0.048 × 30 = 2.88

or v = 1.7 m/s Ans.

Amount by which the spring will be compressed

Let x = Amount by which the spring will be compressed in mm

We know that kinetic energy of the wagon

(^2) 50 (1.7) 2
72.25 kN-m = 72 250 kN-mm
22
mv
== = ...(i)
and compressive load^20
0.05
x
==x
∴ Work done in compressing the spring
= Average load × Displacement
(^20102) kN-mm
2
x
=×=xx ...(ii)
Since the entire kinetic energy of wagon is used to compress the spring, therefore equating
equations (i) and (ii),
72 250 = 10 x^2
∴
2 72 250 7225
10
x ==
or x = 85 mm Ans.
Example 30.19. A bullet of mass 30 g is fired into a body of mass 10 kg, which is suspended
by a string 0.8 m long. Due to this impact, the body swings through an angle 30°. Find the velocity of
the bullet.
Solution. Given : Mass of bullet (m) = 30 g = 0.03 kg and mass of body (M) = 10 kg.
Let u = Initial velocity of the bullet, and
v = Velocity of the body after impact.