Question

Medal!!!


A 1580 kg car is traveling with a speed of 15.0
m / s. What is the net force that is required to bring the car to a halt in a distance of 50.0m?

Answer 1

@hartnn

Answer 2

Sorry it is not a closed question.

Answer 3

use the work/energy principle

work done in stopping the car = energy lost by the car

Answer 4

We're given a formula that has
\[F = mass \times acceleration\]

So we are supposed to use Newton's formula and figure the answer out.

Answer 5

so you'll need the deceleration required to bring the car to the halt

then you can just multiply it with the mass 1580

Answer 6

yes - you can do it that way too

Answer 7

How would Figure out the deceleration? I am only given 15 m / s ?

Answer 8

try this :
\(v^2 = u^2 +2as \)
v = final velocity = 0
u = initial velocity = 15
s = displacement = 50

Answer 9

deceleration = rate of change of velocity

Answer 10

Ok give me a moment :)

Answer 11

So what is "a" in the formula? The acceleration?

Answer 12

yep, the acceleration
and if it comes out to be negative, then deceleration .

Answer 13

Okay

Answer 14

I am a bit confused now:
So I have the equation w/ substitution:
\[0 = 15^2 + 2a50\]

The part where it has 2a50 doesn't make sense. So is it like 2(a + 50)?

Answer 15

\(2\times a\times s \\ 2 \times a \times 50 \\ 100\times a = 100a\)

Answer 16

Oh okay.

Answer 17

\[0 = 15^2 + 100a\]
\[0 = 225 + 100a\]\[-225 = 100a \]
\[a = -2.25?\]

Answer 18

thats right :)

so just multiply it with mass to get the force required..

Answer 19

Wow!!! This was so much easier!

Okay, so it will be:
\[-2.25 \times 1580 = 3555\]

Answer 20

Am I correct?

Answer 21

\(\huge \color{green} \checkmark \)

Answer 22

Thank you!!!!! You're such a great person !! :)

Can you help me with like two more??

Answer 23

Yes or no???

Answer 24

You're a great learner!
and sure :)

Answer 25

Thanks I am creating a new post.