A pile driver hammer of mass 150 kg falls freely through a distance of 5m to strike a pile of mass 400kg and drives it 0.075m into the ground. The hammer does not rebound when driving the pile. Determine the average resistance of the ground.
Thanks in advance.
+37155 mgh = energy of hammer delivered to pile = 150*9.8*5 = 7350 j
Amount delivered to the pile 400kg x 9.8 x .0075=400*9.8*.0075 = 29.4 j
The difference is delivered to the ground (vibrations, ground movement, heat of friction etc) and is the resistance (?)
7350-29.4 = 7320.6 j
(a little unsure)
+2511 Time to test my genetic enhancement.
\(\text {Velocity of hammer at impact with pile}\\ V_f= \sqrt{2ad} \hspace{1em}|\sqrt{2*9.81 m/s^2*5m} = 9.904568m/s\\ \)
\(m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\leftarrow \text {conservation of momentum}\\ 150kg (9.905m/s) = \underbrace{(150 + 400)kg}_{\text{mass of pile & driver}} (v)\leftarrow \text {velocity of combined masses}\\ \hspace {7em} v=\dfrac{150kg (9.905 m/s)}{(150 + 400)kg} = 2.70136m/s\\\)
\(\text {Energies:}\\ \text{kinetic energy} = \frac{1}{2}(550 kg)(0 - 2.70136 m/s)^2 = - 2006.77 J\\ \text{potential energy} = (550 kg) (9.81 m/s^2)(0-0.075 m ) = -404.66 J \\\)
\(\text {Energy to work formula:}\\ \text{(-F)}*(0.075 m) = -2411.43 J \hspace{1em} | (-F) = \text{negative Force} \\ \text{(-F)}=\dfrac{-2411.43J}{0.075m}= 32152.40 \ newtons\\\)
\(\text{Average resistance of the ground }= 32,152.40 \ newtons\)
No figs. Just newtons on this ground.
This concludes the test of my genetic enhancement.
I think I passed.
