Chapter 30 : Work, Power and Energy 619
Now substituting the value of (v^2 = 2gh) in the above equation,(^22)
()
2( )
mgh
E mMgx
mM
×
=++
- 2
()
()
mgh
mMgx
mM
=++ - ...(iv)
We know that work done by the soil resistance
= Rx ...(v)
Since the total energy of the hammer and pile is used in the work done by the soil resistance,
therefore equating equations (iv) and (v),
2
()
()
mgh
mMgxRx
mM
=++=
∴
2
()
()
mgh
R mMg
xm M
=++
Note. Sometimes the pile is of negligible mass. In such cases, the soil resistance,
2
1
mgh h
Rmgmg
xm x
⎛⎞
=+= +⎜⎟
⎝⎠
Example. 30.20. A pile of negligible mass is driven by a hammer of mass 200 kg. If the pile
is driven 500 mm into the ground, when the hammer falls from a height of 4 metres, find the average
force of resistance of the ground.
Solution. Given : Mass of hammer (m) = 200 kg ; Distance through which the pile is driven
into ground (x) = 500 mm = 0.5 m and the height through which hammer falls (h) = 4 m
We know that average force of resistance of the ground,
4
1 200 9.8 1 1960 9 N
0.5
h
Rmg
x
⎛⎞ ⎛ ⎞
=+=× +=×⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠
= 17 640 N = 17.64 kN Ans.
Example 30.21. A hammer of mass 0.5 kg hits a nail of 25 g with a velocity of 5 m/s and
drives it into a fixed wooden block by 25 mm. Find the resistance offered by the wooden block.
Solution. Given : Mass of hammer (m) = 0.5 kg ; Mass of nail (M) = 25 g = 0.025 kg ; Velocity
of hammer (v) = 5 m/s and distance through which nail is driven into wooden block
(x) = 25 mm = 0.025 m.
Let h = Height through which the pile hammer
fell before striking the pile.
We know that velocity of hammer (v),
5 ==×= 2 gh 2 9.8h 19.6h
∴
(5)^225
1.28 m
19.6 19.6
h===
and resistance offered by the wooden block,
2
()
()
mgh
R mMg
xm M
=++
(0.5)^2 9.8 1.28
(0.5 0.025) 9.8 N
0.025 (0.5 0.025)
××
=++
= 238.9 + 5.1 = 244 N Ans.
Fig. 30.12.