Magnitude of a net force?

  • Thread starter rwofford
  • Start date
  • #1
22
0
I have these two problems and I cant seem to find the right answer...

1)A person with a blackbelt in karate has a fist that has a mass of 0.43 kg. Starting from rest, this fist attains a velocity of 5.9 m/s in 0.25 s. What is the magnitude of the average net force applied to the fist to achieve this level of performance?

2)When a 58 g tennis ball is served, it accelerates from rest to a constant speed of 43 m/s. The impact with the racket gives the ball a constant acceleration over a distance of 39 cm. What is the magnitude of the net force acting on the ball?

I know the answers are in N, but i've tried using F=ma which hasn't worked...maybe i am confused about what the magnitude of a net force is...please help!
 

Answers and Replies

  • #2
2,076
2
Show us what you've done and we would able to help you from there.
 
  • #3
399
0
You need to use a very important theorems here: the impulse-momentm theorem

Impulse = average net force * time of application = net change in momentum
 
  • #4
22
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well...i dre the fre body diagram for the first one...and labeled what i know. because i have the initial velocity and elocity and time i solved for the acceleration using: V=Vo+at and I found the acceleration to be 23.6 m/s^2.
For the mass i know it is .43 kg
I used F=ma and i got 10.15 N...
 
  • #5
2,076
2
That seems to be correct, although, technically, you shouldn't be using v = u + at, since it hasn't be explicitly mentioned that the accelaration is constant (in the first problem).

[tex]<a> = \frac{\Delta v}{\Delta t}[/tex], where <a> is the average accelaration.
 
  • #6
22
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so on the second one, i did the same thing only i solved for acceleration by using v^2=Vo^2+2ax and found the acceleration to be 23.7. I used f=ma again and got it to be 1.37...does that sound right?
 
  • #7
2,076
2
Yes, it does.
 
  • #8
399
0
It is not correct to use kinematical equation for constant acceleration here. Impulse-momentum gives the answer in a more technially correct form, and you have to less calculation.
 

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