Newton's laws

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"a person drives their 1452 kg car along a straight road at a constant velocity of 30 m/s east. their brakes suddenly give out. they put the car in neutral and let is coast for 25 seconds. the air drag decelerates the car to a velocity of 25 m/s east. (assume a frictionless surface)."

b) assume the average acceleration while the car is decelerating

so it's just (25 m/s - 30 m/s)/25 s = -0.2 m/s^2?

c) determine the average force of air against the car
f = ma
= (1452 kg)(-0.2)
= 290.4 N

d) after coasting for 25s. they pull their car handbrake to slow the car to a stop. this take 3 seconds. what is the force applied by the handbrake? assume that the force exerted by the air remains constant and is equal to the forced determined in part (c).
so i assume i find acceleration by (0 - 25)/time

but for the time is it (25s-3s) or (3s - 25s)?

and then i just the acceleration times the mass of car = handbrake force?

any help will be appreciated!

~Amy
 

Answers and Replies

  • #2
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You have to consider only the last 3 seconds during which the car is moving.

acceleration times the mass of car = handbrake force?
That's right.
 
  • #3
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thanks neutrino.

for d) i got -8.33 m/s^2 for the acceleration abd -12100N for the force of the handbrake.

in e) it asks "what is the total displacement from the time the brakes give out to the time the car stops"

so for this i used the formula:
d = (0.5)at^2 + v1t
= (o.5)(-8.33)(3^2) + (25m/s)(25s)
= 587.52m

can someone tell me if this is accurate?

~Amy
 
  • #4
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another question i want to check:

"peter and john are playing tug of war on a frictionless surface. peter weighs 539N and john weights 393N. john accelerates towards peter at 3 m/s^2. "

a) what is the magnitude of the force that peter exerts on john?

539N times 3 m/s^2
= 1617N?

any help is appreciated!
~Amy
 
  • #5
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for this one:

"peter and john are playing tug of war on a frictionless surface. peter weighs 539N and john weights 393N. john accelerates towards peter at 3 m/s^2. "

a) what is the magnitude of the force that peter exerts on john?

first i calculated john and peters mass by diving this newton weight by the force of gravity (9.8 m/s^2). so peter is 55 kg, and john is 40 kg.

then for part a) i took peters mass (55 kg) times (-3 m/s^2)
= -165N.

not sure if this is accurate. help?

~Amy
 
  • #6
Doc Al
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physicsgal said:
and then i just the acceleration times the mass of car = handbrake force?
No. Don't forget that the air is still exerting its force.

for d) i got -8.33 m/s^2 for the acceleration abd -12100N for the force of the handbrake.
No, you found the net force.

in e) it asks "what is the total displacement from the time the brakes give out to the time the car stops"

so for this i used the formula:
d = (0.5)at^2 + v1t
= (o.5)(-8.33)(3^2) + (25m/s)(25s)
= 587.52m

can someone tell me if this is accurate?
Divide the motion into two segments and find the displacement in each:
(1) air drag only (coasting)
(2) air drag + hand brake
 
  • #7
Doc Al
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physicsgal said:
for this one:

"peter and john are playing tug of war on a frictionless surface. peter weighs 539N and john weights 393N. john accelerates towards peter at 3 m/s^2. "

a) what is the magnitude of the force that peter exerts on john?

first i calculated john and peters mass by diving this newton weight by the force of gravity (9.8 m/s^2). so peter is 55 kg, and john is 40 kg.
OK.

then for part a) i took peters mass (55 kg) times (-3 m/s^2)
= -165N.
No. To find the force on John, multiply John's mass times John's acceleration.
 
  • #8
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Quote:
for d) i got -8.33 m/s^2 for the acceleration abd -12100N for the force of the handbrake.

No, you found the net force.
so i'd be -121000N - -290.4N?

Divide the motion into two segments and find the displacement in each:
(1) air drag only (coasting)
(2) air drag + hand brake
will do (and will post my results later)

No. To find the force on John, multiply John's mass times John's acceleration.
:blushing: thanks!
so it'd be 120N

what then how would i find the magnitude of the force that john exerts on peter? would it be -120N?

thank you

~Amy
 
  • #9
Doc Al
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physicsgal said:
so i'd be -121000N - -290.4N?
Yes.



so it'd be 120N
Right.

what then how would i find the magnitude of the force that john exerts on peter? would it be -120N?
Yes. Newton's 3rd law should tell you that they exert equal and opposite force on each other. (But realize that magnitudes are always positive. :wink: )
 
  • #10
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in e) it asks "what is the total displacement from the time the brakes give out to the time the car stops"
Divide the motion into two segments and find the displacement in each:
(1) air drag only (coasting)
(2) air drag + hand brake
ok. for (1) i got (0.5)(30m/s + 25 m/s)(25s)
= 687.5 m

for (2) i got (0.5)(-8.33 m/s^2)(3^2)
= -37.485

so 687.5 + - 37.485
= 650.02 m

:biggrin:

more on the john/peter saga..
"sarah decides to join the game as well. now peter pulls on sarah with a force of 45 N east, and john pulls on her with a force of 25N north. what is sarah's resultant acceleration, if she weighs 294N?"
so first of all she weighs 30kg

and to find the c^2 it's (45^2) + (25^2)
= 51.48 N

and to find the tan angle is 26 degree north/west.

so to find sarah's acceleration i go 51.48N/30 kg?

~Amy
 
  • #11
Doc Al
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physicsgal said:
ok. for (1) i got (0.5)(30m/s + 25 m/s)(25s)
= 687.5 m
Good.
for (2) i got (0.5)(-8.33 m/s^2)(3^2)
= -37.485
Nope. Sanity check: It slows down, but doesn't go backwards. :eek:

Use the full equation:
[tex]d = v_0 t + (1/2) a t^2[/tex]
(don't forget the initial velocity!)

You can also use the exact same equation you used for segment (1). :wink:


more on the john/peter saga..

so first of all she weighs 30kg

and to find the c^2 it's (45^2) + (25^2)
= 51.48 N
Good.

and to find the tan angle is 26 degree north/west.
Check this one. Her acceleration should be north-east.

so to find sarah's acceleration i go 51.48N/30 kg?
Good.
 
  • #12
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for (2) i got (0.5)(-8.33 m/s^2)(3^2)
= -37.485

Nope. Sanity check: It slows down, but doesn't go backwards.

Use the full equation:

(don't forget the initial velocity!)
(i knew the negative displacement sounded a bit suspicious, lol)

ok so (25m/s)(3s) + (0.5)(-8.333)(3^2)
= 37.515

650.02+37.515
= 687.54 m :biggrin:

thanks for the help!

~Amy
 
  • #13
Doc Al
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physicsgal said:
(i knew the negative displacement sounded a bit suspicious, lol)

ok so (25m/s)(3s) + (0.5)(-8.333)(3^2)
= 37.515
Good. But you can also use the same equation that you used for segment (1): d = (0.5)(25 m/s + 0 m/s)(3 s)


650.02+37.515
= 687.54 m
I think you copied the answer from (1) wrong.

thanks for the help!
My pleasure.
 
  • #14
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650.02+37.515
= 687.54 m


I think you copied the answer from (1) wrong.
thanks. i just noticed that too. so the total displacement should be 725.02 m east :smile:

~Amy
 

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