Work done by the hammer in a single stroke?

In summary, the conversation discusses the energy and work involved in driving a 9 inch nail into a log with a 10kg hammer. The initial energy of the hammer is calculated as 100J, and it is assumed that 20% of this energy is dissipated during the collision with the nail. This leaves 80J of energy delivered to the nail in a single stroke, assuming the hammer remains at rest after hitting the nail. The work done by the hammer in a single stroke is also assumed to be 100J. However, it is unclear how far into the log the nail is driven in a single stroke, and the participants are seeking hints on how to solve for this information.
  • #1
fro
23
0
Need some verification.

I am trying to drive a 9 inch nail into a log with a 10kg hammer, raising it to a maximum height of 1m and then gain momentum as it is swung down.

a. Energy of the hammer immediately prior to striking the nail:
[tex]10kg\times10\frac{m}{s^2}\times1m = 100J.[/tex]

b. If 20% of hammer's energy is dissipated during the collision, how much energy is delivered to the nail in a single stroke provided the hammer remains at rest after hitting the nail?

[tex]80\%\times100J = 80J.[/tex]

c. Work done by the hammer in a single stroke?

Shouldn't this be the 100J also?

d. How far into the log is the nail driven in a single stroke?

Not sure about this one!

Could someone please give me some hints on these. Thanks.
 
Physics news on Phys.org
  • #2
Did you state all the information you were given for this problem? It seems that some assumption is being made that I am not seeing, and since nobody else has replied I assume others are not seeing it either.
 
  • #3


a. The energy of the hammer immediately prior to striking the nail is 100J.

b. Since 20% of the hammer's energy is dissipated during the collision, only 80J of energy is delivered to the nail. This means that the remaining 20J of energy is lost as heat or sound.

c. The work done by the hammer in a single stroke is equal to the energy delivered to the nail, which is 80J.

d. The distance the nail is driven into the log in a single stroke depends on the resistance of the log and the force exerted by the hammer. This can be calculated using the formula W = Fd, where W is the work done, F is the force, and d is the distance. So, if we assume that the force exerted by the hammer is equal to its weight (98N), then the distance the nail is driven into the log can be calculated as d = W/F = 80J/98N = 0.82m. However, this calculation is only an estimate and the actual distance may vary depending on the specific situation.
 

FAQ: Work done by the hammer in a single stroke?

What is work done by the hammer in a single stroke?

The work done by the hammer in a single stroke refers to the energy transferred from the hammer to the object it strikes. It is a measure of the force applied by the hammer over a certain distance.

How is work done by the hammer calculated?

The work done by the hammer can be calculated by multiplying the force applied by the hammer by the distance it travels. This is known as the work-energy principle and is represented by the equation W = Fd.

What factors affect the work done by the hammer?

The work done by the hammer can be affected by several factors, including the mass of the hammer, the speed at which it strikes the object, and the hardness of both the hammer and the object. These factors can impact the force and distance traveled, ultimately affecting the work done.

How does the work done by the hammer affect the object it strikes?

The work done by the hammer can change the kinetic energy and motion of the object it strikes. The greater the work done, the more energy is transferred to the object, resulting in a greater impact and potential damage.

Can the work done by the hammer be negative?

Yes, the work done by the hammer can be negative if the hammer is pulled away from the object or if the force applied is in the opposite direction of the hammer's motion. In this case, the hammer is doing work to move away from the object, rather than transferring energy to it.

Similar threads

Force applied: calculating increase in temperature (work mostly done, )
Replies
2
Views
5K

Hot Threads

Recent Insights

Back
Top