Trouble with Mechanics Problem

[Tyrzone]

A pile driver of a mass of 800kg falls freely through a height of 2m onto a pile of mass 300kg and drives the pile 350mm into the ground. Determine the resisting force in the ground, assuming that the pile driver remains in contact with the pile and does not bounce.

Firstly I thought I should work out the force and to do that I needed the acceleration so I rearranged a suvat equation which is v²=u²+2as into a= (v²-u²)/2s

After that I thought that there was no initial or final velocity so they would both be 0. That then worked out to be 4m/s.

I worked out force by doing 800x4=3200N and got stuck from there on. Any help from someone who knows where to go next? Thanks

 

[MarkFL]

I would use energy considerations here...what is the initial gravitational potential energy of the pile driver/pile system?

 

[I like Serena]

A pile driver of a mass of 800kg falls freely through a height of 2m onto a pile of mass 300kg and drives the pile 350mm into the ground. Determine the resisting force in the ground, assuming that the pile driver remains in contact with the pile and does not bounce.

Firstly I thought I should work out the force and to do that I needed the acceleration so I rearranged a suvat equation which is v²=u²+2as into a= (v²-u²)/2s

After that I thought that there was no initial or final velocity so they would both be 0. That then worked out to be 4m/s.

I worked out force by doing 800x4=3200N and got stuck from there on. Any help from someone who knows where to go next? Thanks

Hi Tyrzone! Welcome to MHB! ;)

I suggest to use conservation of energy and the definition of work.

The work $W$ done by the ground is given by:
$$W = F_{ground} \cdot d$$
where $d$ is the distance that the pile was driven into the ground.

What is the gravitational energy of the pile driver and the pile, before and after the pile driving?

 

[Tyrzone]

Hi Tyrzone! Welcome to MHB! ;)

I suggest to use conservation of energy and the definition of work.

The work $W$ done by the ground is given by:
$$W = F_{ground} \cdot d$$
where $d$ is the distance that the pile was driven into the ground.

What is the gravitational energy of the pile driver and the pile, before and after the pile driving?

Thanks for the helps. I think I got the right answer but I thought I would check here first. Would it be: 17.93kN/17930/17.93x10³?

 

[MarkFL]

I find the initial energy is:

\(\displaystyle E_i=\left(800\cdot9.8\cdot2.35+300\cdot9.8\cdot0.35\right)\text{ J}=19465\text{ J}\)

Since the final energy is zero, this energy must have been dissipated during the time the ground was working to stop the system. Using the work-energy theorem and the relationship between work and the distance over which a force is exerted, we then find:

\(\displaystyle \overline{F}=\frac{W}{d}=\frac{19465\text{ J}}{0.35\text{ m}}=55580\text{ N}\)

 

[Tyrzone]

I find the initial energy is:

\(\displaystyle E_i=\left(800\cdot9.8\cdot2.35+300\cdot9.8\cdot0.35\right)\text{ J}=19465\text{ J}\)

Since the final energy is zero, this energy must have been dissipated during the time the ground was working to stop the system. Using the work-energy theorem and the relationship between work and the distance over which a force is exerted, we then find:

\(\displaystyle \overline{F}=\frac{W}{d}=\frac{19465\text{ J}}{0.35\text{ m}}=55580\text{ N}\)

Wouldn't it be (1100x9.8x0.35)+(1/2x1100x4.55)=6275.5 due to kinetic energy once the pile has been hit

6275.5/0.35=17930N

 

[I like Serena]

Wouldn't it be (1100x9.8x0.35)+(1/2x1100x4.55)=6275.5 due to kinetic energy once the pile has been hit

6275.5/0.35=17930N

It would help if you show your workings... (Wasntme)

For reference, the kinetic energy of the combined system, right after impact is:
$$E_{combined} = \frac 12 (m_{driver} + m_{pile}) v_{combined}^2=m_{driver}\cdot g\cdot s $$
So:
$$v_{combined}^2=\frac{2\, m_{driver}\cdot g\cdot s}{m_{driver} + m_{pile}}
= \frac{2 \cdot 800\text{ kg}\cdot 9.81 \frac{\text{m}}{\text{s}^2} \cdot 2\text{ m}}{800\text{ kg} + 300 \text{ kg}} = 28.5 \frac{\text{m}^2}{\text{s}^2}
$$

How did you get 4.55? (Wondering)