(^614) „„„„„ A Textbook of Engineering Mechanics
Now substituting value of (P = m.a) in equation (i),
W = ma × s = mas ...(ii)
We know that v^2 = u^2 – 2 as ...(Minus sign due to retardation)
∴ 2 as = u^2 ...(Q v = 0)
or
2
2
u
as=
Now substituting the value of (a.s) in equation (ii) and replacing work done with kinetic energy,
2
KE
2
mu

Cor. In most of the cases, the initial velocity is taken as v (instead of u), therefore kinetic
energy,
2
KE
2
mv

Example 30.17. A truck of mass 15 tonnes travelling at 1.6 m/s impacts with a buffer
spring, which compresses 1.25 mm per kN. Find the maximum compression of the spring.
Solution. Given : Mass of the truck (m) = 15 t ; Velocity of the truck (v) = 1.6 m/s and buffer
spring constant (k) = 1.25 mm/ kN
Let x = Maximum compression of the spring in mm.
We know that kinetic energy of the truck
(^2) 15 (1.6)^2
19.2 19 200 kN-mm
22
mv
== == ...(i)
and compressive load 0.8 kN
1.25
x
==x
∴ Work done in compressing the spring
= Average compressive load × Displacement
0.8 0.4 (^2) kN-mm
2
x
=×=xx ...(ii)
Since the entire kinetic energy of the truck is used to compress the spring therefore equating
equations (i) and (ii),
19 200 = 0.4 x^2
∴
2 19 200 48 000
0.4
x ==
or x=219 mm Ans.
Example 30.18. A wagon of mass 50 tonnes, starts from rest and travels 30 metres down a
1% grade and strikes a post with bumper spring as shown in Fig. 30.8.
Fig. 30.8.
If the rolling resistance of the track is 50 N/t, find the velocity with which the wagon strikes
the post. Also find the amount by which the spring will be compressed, if the bumper spring compreses
1 mm per 20 kN force.