Average Retarding Force Problem

  • Thread starter Thread starter thebestrc
  • Start date Start date
  • Tags Tags
    Average Force
AI Thread Summary
An arrow with a mass of 45 g strikes a wood block at 70 m/s and penetrates 280 mm. The time taken for the arrow to stop is calculated to be approximately 0.008 seconds, correcting an earlier miscalculation of 7.14 seconds. The average retarding force exerted by the wood block is determined to be around -393.75 N, after clarifying that the acceleration should not include the effects of gravity. The discussion highlights the importance of focusing on the specific motion involved and not getting sidetracked by unnecessary calculations. Overall, participants emphasize the learning process and overcoming challenges in understanding physics concepts.
thebestrc
Messages
10
Reaction score
0
[SOLVED] Average Retarding Force Problem

Homework Statement



An arrow of mass 45 g strikes a wood block target horizontally with a velocity of 70
ms-1 and penetrates a distance of 280 mm. Ignoring air resistance, calculate (a) the time it takes for the arrow to stop moving, and (b) the average retarding force exerted on the arrow by the wood block target.

Homework Equations



v=u + at
F= ma
V2=u + 2as

The Attempt at a Solution



I worked out the time using the 1st formula listed (t=7.14 3sf) but when i go to work out the average retarding force using F=ma and get 0.44145 which seems wrong to me .Could someone please tell where I'm going wrong?
I also worked out the total distance traveled by the arrow which is 250m including the 280mm penetration of the wood.
 
Physics news on Phys.org
For the first part I'd use the following since 7.14 seems a lot to me.

s = \left(\frac{u+v}{2} \right) t

Once you have time you can work out the acceleration and then the force.
 
I've worked out the total distance traveled by the arrow which is 250m . I found this out using the equation v^2=u^2 +2as . I then substitute it into the equation you gave me and i still get 7.14 (3sf).

u=70 , a=-9.81 s= 250 (including distance penetrated)
 
It doesn't ask for the the time it takes for the arrow to travel the full distance. It asks how long it takes to stop moving. That is how long it takes to come to rest in 280mm.
 
oh ok. me thinks i interpretated the question. Thanks.
that means t=0.008s.
I thought a= -9.81m/s since the effects of air/wind resistance are ignored?

I think this is the right answer for retarding force: -393.75N
I got it by working out a instead of using -9.81 , and i got -8750 which seems quite alot.
 
Last edited:
-9.81 will be the vertical component of acceleration but we're only considering the horizontal motion. Now you have your time you can work out the acceleration the arrow underwent to cause it to stop. Then you can work out the force.

Actually I've been a bit of a fool because you don't really need to work out the time. Its just since you said you'd worked it out in the first post I kind of got blinkered by it. Of course you could have worked out the acceleration directly from v^2 = u^2 +2as. Never mind. You get the same answer in the end. :smile:
 
Thanks for the help. I've been at uni for a year now and this stuff still puzzles me.
 
thebestrc said:
Thanks for the help. I've been at uni for a year now and this stuff still puzzles me.

You'll get there. I was pretty much puzzled initially because before that I'd never really worked at my education. Once I realized I had to start putting in some work it took a while to learn how to learn but I made it. As will yourself.
 
Back
Top