Impulse: Finding the average force

[impstudies]

Homework Statement

The 2kg head of an axe is moving at 60m/sec when it strikes a log. If the blade of an axe penetrates a 20mm into the log, find the average force it exerts.

m= 2kg
V= 60m/sec
d= 20mm or 0.02m

Homework Equations

F= pFINAL - pINITIAL / time
t= d/v

The Attempt at a Solution

m1v1 = axe
m2v2 = log

t = 0.02/60
= 3.33sec
F = [m1v1 + m2v2] - [m1v1 + m2v2] / time
= [(2kg)(60m/sec)+(m2)(0)] - [(2kg)(m2)(0)] / time
= (120 kg.m/sec + m2) / time ?

^ considering the log has 0 velocity? but has no mass given?

I solved for the time then used the impulse formula pFINAL-pINITIAL/time but since there is no mass of the log given I'm baffled because I feel like I'm using the wrong equations. I was lost if I should find the momentum or impulse since the axe makes contact with the log. I'm really bad at physics, your help is so appreciated :(

 

[Doc Al]

t = 0.02/60
= 3.33sec

Don't assume the speed is constant. The axe comes to rest. (What's the average speed of the axe?)

F = [m1v1 + m2v2] - [m1v1 + m2v2] / time
= [(2kg)(60m/sec)+(m2)(0)] - [(2kg)(m2)(0)] / time
= (120 kg.m/sec + m2) / time ?

^ considering the log has 0 velocity? but has no mass given?

Forget about the log. Just analyze the change in momentum of the axe.

 

[impstudies]

i assume the speed here is the velocity?
the axe comes to rest so its Vf = 0, with Vi = 60 m/sec

= vf-vi
= 0-60
= -60m/sec

is that right?

 

[Doc Al]

i assume the speed here is the velocity?
the axe comes to rest so its Vf = 0, with Vi = 60 m/sec

= vf-vi
= 0-60
= -60m/sec

is that right?

You found the change in speed. If you want to calculate the time for the axe to come to rest, you need the average speed.

 

[rwisz]

Hello, thank you for posing this question. In order to find the average force exerted by the axe on the log, we need to use the equation F = m*a, where m is the mass of the object and a is the acceleration. In this case, the acceleration can be found using the equation a = v/t, where v is the velocity and t is the time.

First, we need to convert the distance of penetration from millimeters to meters, so 20mm becomes 0.02m. Then, we can use the equation t = d/v to find the time it takes for the axe to penetrate the log, which is 0.02m / 60m/sec = 0.00033sec.

Next, we can use the equation a = v/t to find the acceleration of the axe as it penetrates the log, which is 60m/sec / 0.00033sec = 181818.18 m/s^2.

Now, we can plug in the values into the equation F = m*a, where m is the mass of the axe, which is given as 2kg. Therefore, the average force exerted by the axe on the log is F = (2kg)(181818.18 m/s^2) = 363636.36 N.

I hope this helps clarify the process of finding the average force exerted by the axe on the log. Remember, it is important to always use the correct equations and units when solving problems in physics. Keep practicing and seeking help when needed, and you will improve your understanding of the subject.