A pile driver drives a post into the ground. The mass of the pile driver is 2500 kg and it is dropped through a height of 8.0 m on each stroke. If the resisting force of the ground is \( 4.0^{\circ} 10^{6} \mathrm{~N} \), how far is the post driven in on each stroke? 1) 4.9 cm 2) 9.8 cm 3) 16 cm 4) 49 cm 5) 54 cm

To find out how far the post is driven in on each stroke, we can use the energy principle. The potential energy (PE) of the pile driver when dropped can be calculated with the equation: \[ PE = mgh = (2500 \, \text{kg})(9.81 \, \text{m/s}^2)(8.0 \, \text{m}) = 196,200 \, \text{J} \] The work done against the resisting force (W) can be described by: \[ W = F \cdot d \quad \text{(where \(F\) is the resisting force and \(d\) is the distance driven)} \] Setting the potential energy equal to the work done, \[ 196,200 \, \text{J} = (4.0 \times 10^6 \, \text{N}) \cdot d \] Solving for \(d\): \[ d = \frac{196,200 \, \text{J}}{4.0 \times 10^6 \, \text{N}} \approx 0.04905 \, \text{m} \approx 4.9 \, \text{cm} \] So the answer is **1) 4.9 cm**.