Net Force Questions

1. Oct 19, 2005

thua

1. A 1.2 x 10^4 kg truck is travelling south at 22 m/s.
a) What net force is required to bring the truck to a stop in 330 m?
b) What is the cause of this net force?
Answer: a) 8.0 x 10^3 N [N]

2. Each of the four wheels of a car pushes on the road with a force of 4.0 x 10^3 N [down]. The driving force on the car is 8.0 x 10^3 N [W]. The frictional resistance on the car is 6.0 x 10^3 N [E]. Calculate the following:
a) the mass of the car
b) the car's acceleration
Answer: a) 1.6 x 10^3 kg b) 1.2 m/s^2 [W]

I have the answers to these two questions, but I really need help on how to understand and do it.

So far...
1. a) v = d/t
t = d/v
t = (0.33 km)/(22 m/s )
t = 0.015 s

a = v/t
a = (22 m/s )/(0.015 s)
a = 1466.7 m/s^2 ?? <====== stopped here because this makes no sense..

Last edited: Oct 19, 2005
2. Oct 19, 2005

Seiya

the time makes no sense... you are right :p

t = (0.33 km)/(22 m/s )

the distance has to be in meters...

3. Oct 19, 2005

whozum

If you've covered energy then that would be the best way to solve the problem. If not then the following equation should look familiar:

$v_f^2 = v_i^2 + 2ad$ which is what you should use to tackle this problem.

4. Oct 19, 2005

thua

whozum... is that to find the acceleration?

(22 m/s )^2 = (0 m/s )^2 + 2a(330 m)
a = 0.73 m/s^2

f = ma
f = 8760 N <----- something's still wrong

Last edited: Oct 19, 2005
5. Oct 19, 2005

Seiya

if you want the truck to stop acceleration has to be negative...

6. Oct 19, 2005

thua

that still doesn't work though because the net force comes up to -8670 N .

i think i'm totally missing out on something here...

7. Oct 19, 2005

whozum

The acceleration is not south but north, -8670 N = 8670N [N]. However I don't know why they have 8.0 x 10^3N as the answer, your answer of 8670 is alomst correct.

By the way the correct solution is

$v_f^2 = v_i^2 + 2ad$ you had the velocities mixed up.

$(0)^2 = (22)^2 + 2a(330m)$

$a = \frac{22^2}{-660} = -0.733333 [/tex] [itex] F = ma = m(-0.7333333) = 8800N = 8.8 \times 10^3 N [N]$

Last edited: Oct 19, 2005
8. Oct 19, 2005

thua

okay thankssss... and for number 2... i really dont know how itll work cause mass is equal to net force over acceleration, but all i have is net force... no acceleration.... so how would that work out?

9. Oct 19, 2005

whozum

The force from the 4 tires is the weight of the car (normal force). You can find the mass from there.

10. Oct 19, 2005

thua

does that mean that the answer is wrong again? because i came up with 16000 N which is 1.6 x 10^4 and they say its 1.6 x 10^3 kg.

and does 1 N equal to 1 kg.... im not very familiar with that.
but 1 N actually equals to 1 kg x m/s^2, though right?

Last edited: Oct 19, 2005
11. Oct 19, 2005

Seiya

you answered your own question a newton is a kg * m/s^2