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Net Force Questions

  • Thread starter thua
  • Start date
9
0
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:
43
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the time makes no sense... you are right :p

here is your problem

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

the distance has to be in meters...
 
2,208
1
thua said:
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]
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:

[itex] v_f^2 = v_i^2 + 2ad [/itex] which is what you should use to tackle this problem.
 
9
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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
 
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43
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if you want the truck to stop acceleration has to be negative...
 
9
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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...
 
2,208
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thua said:
that still doesn't work though because the net force comes up to -8670 N .

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

[itex] v_f^2 = v_i^2 + 2ad [/itex] you had the velocities mixed up.

[itex] (0)^2 = (22)^2 + 2a(330m) [/itex]

[itex] a = \frac{22^2}{-660} = -0.733333 [/tex]

[itex] F = ma = m(-0.7333333) = 8800N = 8.8 \times 10^3 N [N] [/itex]
 
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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?
 
2,208
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The force from the 4 tires is the weight of the car (normal force). You can find the mass from there.
 
9
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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:
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you answered your own question a newton is a kg * m/s^2
 

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