Question

18. A 2.0-gram bullet is shot into a tree stump. It enters at a speed of 3.00 x 104 cm/s and comes to rest after having penetrated 0.05 m in to the stump. What was the average force during the impact?

Answer 1

@TheSmartOne

Answer 2

I don't know. Sorry :/

maybe these users might be able to help you :)
@abb0t @Abhisar @Michele_Laino @ParthKohli @iambatman

Answer 3

now i must play the waiting game, thanks for you the suggestions.

Answer 4

Haha, np :)

Answer 5

\[\rm W = \Delta KE = F_{avg} \times distance\]

Answer 6

I tried to put it in simple mathematical symbols. Do tell if you need further help.

Answer 7

I'm in a pretty deep hole with this. I'm not sure where the numbers are plugged in.

Answer 8

I would have to assume that: W = work; KE = kinetic energy; F = force but what does the triangle mean?

Answer 9

OK, let's first start with the meaning of those symbols.

In stopping the bullet, the tree is doing some work through reactionary force, friction and stuff. When work is done, there is always change in energy. Is it possible to track the change in that energy? It definitely is. The bullet possesses some kinetic energy as it goes into the tree. As it begins to penetrate, it loses the kinetic energy (remember - kinetic energy is the energy of motion). What exactly, in numerical terms, is this change? \(\)We can represent change in KE as \(\\\rm \Delta KE\), where \(\Delta\) (delta) is a symbol standing for "change".

It is also a fact that work done through a distance is always force applied times that distance (displacement, actually). And this displacement has been given as 0.05 metres.

Answer 10

\[\rm \Delta KE \equiv KE_{final} - KE_{initial}\]

Answer 11

parthkohli is correct and because the bullet goes to rest the Vfinal is 0 meaning you have only the initial kinetic energy and you are just solving for force after.

Answer 12

so KE final = 0

Answer 13

how do i solve for the initial KE?

Answer 14

KE = .5*m*v^2 you have all everything you need from the problem itself.

Answer 15

I'm still confused, I used up 2 pages in my notebook trying to figure this out. What am i doing wrong?

Answer 16

KEinitial =1/2*0.002kg*300^2m/s is this right?

Answer 17

Yes, that is correct.

Answer 18

but how do i simplify this?

Answer 19

or does it need to be simplified at all?

Answer 20

thats how it gets swaggin...

Answer 21

\[DeltaKE = 0 - 1/2*0.002kg*300^2m/s\]

Answer 22

you simplify it by multiplying the value out and getting a single number then use that back in the work equation to get the average force.

Answer 23

how do i multiply them if there are no like terms? Or does that mathematical rule even apply in physics?

Answer 24

what do you mean? -.5*.002*300^2 = what? the units combine to make kg*m^2/s^2 which is the unit for energy if that is what you are talking about.

Answer 25

after you find that answer use that in W = deltaKE = F * distance .... you now have deltaKE and the distance is stated in the problem as .05m so you have F = (.5*.002*300^2)/.05

Answer 26

(.5*.002*300^2)/5 = (.5 * .002*90000)/5 = 90/5 = 18

Answer 27

so F = 18

Answer 28

I didn't do the calculation but provided you inputed the numbers correctly then yes F = 18 N

Answer 29

so you're saying there's a chance that the problem wasn't done correctly?

Answer 30

no it was done incorrectly because you divide by 5m instead of .05 m the answer should be 1800 N

Answer 31

all your conversions and everything else was right you just had the wrong distance in the last part.

Answer 32

Thank You so much, this problem was giving me a rough day.