What is the force applied by the handbrake to stop the car?

[baztack]

Hi I am getting into the section of Newtons laws and came across this question which I found difficult.

The question reads:

Roberta drives her 1452kg car along a straight, level road at a constant velocity of 30.0 m/s {E}. Her brakes suddenly give out. She puts the car in heutral and let's it coast for 25.0s. The air drag decelerates the car to a velocity of 25.0m/s[E].(Assume a frictionless surface)

A) Draw a freebody diafra, of the car before the breaks give out and while coasting... I think I got this one right...

B)Determine the avergae accleration furing this interval...
My answer: G: deltaTime - 25.0s, deltaVelocity - 30.0m/s
R: acceleration
A: a=deltav/deltat
s: a = 30/25 =1.2m/s
P: avg acceleration is 1.2m/s

C) Determine the average force of air against the car.
G: Mcar=1452kg, accel=1.2m/s
R: Fair
A: Fair=(m)(accel)
S: Fair=1452kg x 1.2m/s
=1742.4N
P: Avg force of air against the car is 174x10 N.
Im assuming up to this point that my calculations are correct... if you see a problem please hint and let me know...

This is where I am running into trouble...
D) After coasting for 25.0s, Roberta pulls her handbrake to slow the car to a stop. If it takes 3.0s to stop the car, what is the force applied by the handbrake? Assume that the force exerted by the air remains constand and is equal to the force determined in part c.

I have my givens mapped out G: m=1452kg, V1=30.0m/s, V2=25.0m/s, T1=25.0s, T2=28.0s, Fair=1742.4N
R: Fapp?

For the rest of the question I was trying to firgure out exactly how to go about finding a solution to this problem. I know that F=m(accel) so should I use an acceleration equation to find the accleration then plug it into the F=M(accel) to get the force of the applied handbrake?

Sorry I've just been stuck on this question now for hours and any help would be greatly appreciated.

 

[Gokul43201]

Roberta drives her 1452kg car along a straight, level road at a constant velocity of 30.0 m/s {E}. Her brakes suddenly give out. She puts the car in heutral and let's it coast for 25.0s. The air drag decelerates the car to a velocity of 25.0m/s[E].

My answer: ... deltaVelocity - 30.0m/s

Recheck that !

 

[VietDao29]

After, you have recheck your answer, and refind the force of air. Here's some hints:
There are two force that slows the car down : air force and (hand brake) friction force.
You have already known the air force. You can also know the acceleration of the car (you know $\Delta t = 3.0 s$ and $\Delta v = - 25.0 m/s$).
You also know the mass of the car. Can you work out the force of friction?
Viet Dao,
EDIT:
Just one more thing : acceleration is $m / s ^ {2}$ not $m / s$ like you type in your post.

 

[cscott]

P: avg acceleration is 1.2m/s

I thing to remember here is that the car in your problem is slowing down, thus the acceleration cannot be positive. Little things like these help you quickly spot simple mistakes.

 

[mukundpa]

The equations you are using are for motion with uniform acceleration only. When the handbrake is applied the force and hence the acceleration is changed, so we have to solve the second part seperately
Vin = 25 m/s ; Vfin = 0; F = -(1742.4 + Fb) and t = 3 s

 

[baztack]

Hi thanks for the input... I realize acceleration is m/s squared but I am not sure how to type the squared sign on the computer...

Also in the original question is says to asssume a frictionless surface so I don't think it is necessary to find the force of friction...

After, you have recheck your answer, and refind the force of air. Here's some hints:
There are two force that slows the car down : air force and (hand brake) friction force.
You have already known the air force. You can also know the acceleration of the car (you know $\Delta t = 3.0 s$ and $\Delta v = - 25.0 m/s$).
You also know the mass of the car. Can you work out the force of friction?
Viet Dao,
EDIT:
Just one more thing : acceleration is $m / s ^ {2}$ not $m / s$ like you type in your post.

 

[shill]

$m / s ^ {2}$

 

[zwtipp05]

Also in the original question is says to asssume a frictionless surface so I don't think it is necessary to find the force of friction...

The hand brake uses friction to slow the car, that is what he is referring to.

 

[baztack]

The hand brake uses friction to slow the car, that is what he is referring to.

Oh okay thank you for clearing that up... I will be finishing my attemp at these questions and a bunch more later tonight/tomorrow morning and will post my conclusions.. thank you all for your help thus far.

 

[baztack]

Okay I tried to solve part b,c and d again...

b) G:DeltaV= -25m/s, DeltaT=25.0s
R: Accel
A: Accel= deltaV/deltaT
s: Accel= -25m/s / 25s = -1.0m/s^2
p: The avg acceleration of the car is -1.0m/s^2 during this interval.

c) G:Accel=-1.0m/s^2, Mcar=1542kg
R: Fair
A: Fair=(m)(accel)
s: Fair=(1542kg)(-1.0m/s^2) = 1542N
p: The average force of the air is -1542N.

d) G: F= -(-1542kg + Fb), Accel=-1.0m/s^2, Mcar=1542kg
R: Fb
A: F=(m)(accel)
S: 1542 - Fb = (1542Kg) (-1.0m/s^2)
-Fb= -1542N - 1542N
-Fb= -3084N
Fb= 3084N
P: The force applied by the handbrake is 3084N


Any comments about these conclusions would be appreciated... I have a strange feeling I am still off in a couple of them.

 

[zwtipp05]

Okay I tried to solve part b,c and d again...

b) G:DeltaV= -25m/s, DeltaT=25.0s
R: Accel
A: Accel= deltaV/deltaT
s: Accel= -25m/s / 25s = -1.0m/s^2
p: The avg acceleration of the car is -1.0m/s^2 during this interval.

c) G:Accel=-1.0m/s^2, Mcar=1542kg
R: Fair
A: Fair=(m)(accel)
s: Fair=(1542kg)(-1.0m/s^2) = 1542N
p: The average force of the air is -1542N.

d) G: F= -(-1542kg + Fb), Accel=-1.0m/s^2, Mcar=1542kg
R: Fb
A: F=(m)(accel)
S: 1542 - Fb = (1542Kg) (-1.0m/s^2)
-Fb= -1542N - 1542N
-Fb= -3084N
Fb= 3084N
P: The force applied by the handbrake is 3084N


Any comments about these conclusions would be appreciated... I have a strange feeling I am still off in a couple of them.

Ok, for part b, you are looking for the average acceleration during coasting.
She goes from 30 m/s to 25 m/s over a time of 25 s. Therefore we get 5 m/s for change in velocity and 25 s for change in time. So a = 5/25 = .2 m/(s^2)

For part c you have the right method, but your answer was wrong due to your error in part b. F=ma F= 1542 kg * .2 m/(s^2)
F = 308.4 N

For part d, you should approach like parts b and c combined.
The car goes from 25 m/s to 0 m/s in 3 s. So change in velocity is 25 m/s and change in time is 3 s.
This gives an acceleration of 25/3 or a=8.33 m/(s^2)

Now use the method in part c.
F= ma
308.4 + Force of the Hand Brake = 1542 kg * 8.33 m/(s^2)

We get 308.4 + Force of the Hand Brake = 12850
So Force of the Hand Brake = 12541.6 N

You have to be very careful and make sure you only looking at what the changes are in the time frame indicated.

BTW, I'm curious, What does RASP stand for?

 

[baztack]

Thank you for your reply... rasp is actually Grasp and its a method which was taught to me to somewhat organize your solutions.

G- givens
R- what's Required
A- The Analysis ie) eqns that will be used
S- The Solution where all the work is done
P- the paraphrase to answer the original question in words.