Answer to Question #154643 in Mechanics | Relativity for Colm

Question

A pile drive hammer of mass 200kg falls freely through a distance of 8.2m to strike a pile of pass 800kg and drives it 105mm into the ground. The hammer does not rebound when driving the pile.

Determine the average resistance of the ground using the principle of conservation of energy.

Answer 1

According to the energy conservation law, the energy given to the pile by the hammer is equal to the initial potential energy of the hammer:

$$W = m_1gh$$

where $m_1 = 200kg$ is the mass of the hammer, $h = 8.2m$ is its initial height, and $g = 9.81m/s^2$ is the gravitational acceleration. Since the hammer does not rebound when driving the pile, all this energy was transmitted into a work against ground resistance:

$$W = A\\ A = Fs$$

where $F$ is the average resistance force, and $s = 105mm = 0.105m$ is the distance travelled by the pile. Thus, obtain:

$$m_1gh = Fs\\ F = \dfrac{m_1gh}{s}\\ F = \dfrac{200\cdot 9.81\cdot 8.2}{0.1} = 160884N$$

Answer. 160884 N.