Question

a 6*10^(-2) arrow hits the target at 25 m/s and penetrates 3.8 cm before stopping. a) what average force did the target exert on the arrow? …

Answer 1

The most neglected kinematic equation by students seems to be\[v_f^2=v_i^2+2ad\]in this case final velocity is zero and acceleration is negative, so we have\[0=v_i^2-2ad\]we can use this to find the acceleration.
We can then find the force from the formula\[F=ma\]

Answer 2

why would your final velocity be zero? would it be 25m/s instead of zero? so, would it be \[25^2 -0^2=2a(0.038m)\]

Answer 3

" hits the target at 25 m/s and penetrates 3.8 cm before stopping."
it starts at 25m/s and comes to a stop

the way I wrote my formula though it winds up positive like yours

Answer 4

ok. i see it now so it would be \[0^2-25^2=2a(0.038)\]
i am assuming we convert 3.8cm to meters so it would be 0.038 meters.
the answer for a will be\[-625=0.076a\]
\[a=-8223 m/s^2\]
i have a feeling that the acceleration is incorrect because it is to big of a number.

Answer 5

Your acceleration looks right, but if the mass is small this is quite feasible:
it would be\[0=v_i^2-2ad\]\[a=\frac{v_i^2}{2d}\]and the force is \[F=ma=m\frac{v_i^2}{2d}=\frac{(0.06)(25)^2}{2(0.038)}=493\text N\]which seems about right for an arrow hitting something at full force

Answer 6

it makes sense now. i almost got confused of the process you did, but it is exactly like mine, but yours is simplified. another question, so in a world problem if they give us velocity in a newton (force) problem, do we have to use the 1D kinematics equation to find a?

Answer 7

Usually you need to find acceleration when force is involved like this if you don't already know it, yes

Answer 8

sometimes you use other formulas to find a, like the formula for centripetal acceleration from rotational motion

Answer 9

...but in general, if you want to connect motion and force you need to find the acceleration by some means

Answer 10

sorry to bug you. but for the second question: What average force did the arrow exert on the target, would it just be -493N because the arrow and the target together hits at the same magnitude(force) but goes in the opposite direction based on the third newton law?

Answer 11

No problem
Yeah it would just have the opposite sign of the acceleration.

Which is positive and which is negative in not important. What is important is that the force of the target on the arrow has the same sign as the acceleration, and that the force of the arrow in the target has the opposite sign as acceleration.

Answer 12

you can see from the drawing (I hope) that the force of the arrow Fa is opposite the direction of the acceleration, and the force of the target Ft is the same direction as the acceleration

Answer 13

here it is with Fa and Ft a different color to make it more clear