# What is the work done by braking force on a decelerating car?

• Lauren0619
In summary, the conversation was about calculating the work done by a braking force to bring a 1480 kg car to a stop from a speed of 91.0 km/hr. The formula used was 1/2(mass)(v^2), with v being converted to m/s. The crucial point discussed was the number of significant figures to use in the calculations, with the final answer being either -4.73 x 10^5 or -4.74 x 10^5. The conversation also emphasized the importance of including the correct units in the final answer.
Lauren0619

## Homework Statement

With brakes fully applied, a 1480 kg car deccelerates from a speed of 91.0 km/hr. What is the work done by the braking force in bringing the car to a stop?[/B]

1/2(mass)(v^2)

## The Attempt at a Solution

1/2(1480)(25^2) = 462500
I converted km/hr to m/sec to get 25 for the v value. They want us to include the sign of the acceleration so i added a negative sign to make the answer -462500 but it was wrong (I also tried it without the negative sign). What am I missing? Thank you so much![/B]

Jamison Lahman
Consider the number of significant figures given in the numerical data. When you calculate v in m/s, you should keep at least that same number of significant figures. (I like to keep an extra significant figure during the intermediate calculations and then round off to the appropriate number of significant figures at the end of the calculation.)

Lauren0619
TSny said:
Consider the number of significant figures given in the numerical data. When you calculate v in m/s, you should keep at least that same number of significant figures. (I like to keep an extra significant figure during the intermediate calculations and then round off to the appropriate number of significant figures at the end of the calculation.)
Thanks for your quick reply, TSny, For the significant figures and multiplication I know that you use the value with the least amount of numbers (precision). So for this one would I be basing it off of .5? or 25? If it was .5 I would write it as 4x10^5. If it was 25, I would write it as 4.6 x 10^5. I'm just not sure which one to use? And I would still include the negative sign for the acceleration, do you agree?

How many significant figures are in the quantity 91.0 km/hr?
When you convert this to m/s, how many significant figures should you retain?

TSny said:
How many significant figures are in the quantity 91.0 km/hr?
When you convert this to m/s, how many significant figures should you retain?
Trailing zeros are significant, so 3 sig figs are in 91.0. So converting to m/s it would be 25.0 m/s? So then my answer should be -4.62 x 10^4 retaining the 3 significant figures? Thanks again for being helpful and patient.

Lauren0619 said:
Trailing zeros are significant, so 3 sig figs are in 91.0.
Yes.
So converting to m/s it would be 25.0 m/s?
You have the correct number of significant figures, but the last digit is not 0. Check your conversion of km/hr to m/s.

Lauren0619 said:
Trailing zeros are significant, so 3 sig figs are in 91.0. So converting to m/s it would be 25.0 m/s? So then my answer should be -4.62 x 10^4 retaining the 3 significant figures? Thanks again for being helpful and patient.
It looks like you already got your answer. To be explicit, the .5 is a scalar therefore its significance is neglected (see https://en.wikipedia.org/wiki/Significance_arithmetic). I agree with your negative notation since you're going from a higher speed to rest, hence decelerating. (For further clarification, see https://www.av8n.com/physics/acceleration.htm).
Cheers

Lauren0619
Check your conversion to m/s -- see this for a step-by-step guide.

Using 25 m/s vs 25.3 m/s makes a huge difference in the answer

Jamison Lahman and Lauren0619
Isabelle Edmond said:
Check your conversion to m/s -- see this for a step-by-step guide.

Using 25 m/s vs 25.3 m/s makes a huge difference in the answer
Thanks everyone! I redid my conversion and got 25.3 (rounding up) for my meters per second. Recalculating I got 4.73 x 10^4. Keeping 4.73 consistent with 3 significant figures. And I will mark it as negative since it is decelerating. Did I finally get it?

Lauren0619 said:
Thanks everyone! I redid my conversion and got 25.3 (rounding up) for my meters per second. Recalculating I got 4.73 x 10^4. Keeping 4.73 consistent with 3 significant figures. And I will mark it as negative since it is decelerating. Did I finally get it?
The 4.73 looks good, but I think you should check the 104 part.

Lauren0619
TSny said:
The 4.73 looks good, but I think you should check the 104 part.
That was a silly mistake on my part. Thanks for catching that. My answer in full is 473666.6 which converted into scientific notation will result in a complete answer of -4.73x10^5 or should I round it up to -4.74x10^5? Again, thanks all for the patience.

If you round 473666.6 to three significant figures, you would "round up" to obtain 474000.
As I mentioned in an earlier post, I prefer to keep an extra significant figure in intermediate calculations.
So, I would have obtained v = 25.28 m/s. This would give -472918 for the work, which rounds to -473000 to three significant figures. You can see how you might obtain either -4.73 x 105 or -4.74 x 105. So, in my opinion, either answer should be acceptable. I leave it to you to include the correct units.

Lauren0619

## 1. How does the deceleration of a car affect its stopping distance?

The deceleration of a car is directly proportional to its stopping distance. This means that the greater the deceleration, the shorter the stopping distance will be. This is because a higher deceleration means the car is slowing down at a faster rate, which allows it to come to a complete stop sooner.

## 2. What factors contribute to the deceleration of a car?

There are several factors that can contribute to the deceleration of a car, including the braking force applied by the driver, the mass of the car, the condition of the brakes, and the surface friction of the road. Other external factors, such as air resistance and incline of the road, can also affect the deceleration of a car.

## 3. How does the speed of a car impact its deceleration?

The speed of a car has a significant impact on its deceleration. The higher the speed, the longer it will take for the car to come to a complete stop. This is because at higher speeds, the car has more kinetic energy, which needs to be dissipated through braking. Additionally, the faster a car is traveling, the longer the reaction time for the driver to apply the brakes, which can also affect the deceleration.

## 4. How do different road conditions affect the deceleration of a car?

The condition of the road surface can greatly impact the deceleration of a car. For example, a wet or icy road will have less friction, making it more difficult for the car to slow down. This can result in a longer stopping distance and potentially lead to skidding or loss of control. On the other hand, a dry road with good traction will allow for better deceleration and shorter stopping distances.

## 5. Can the deceleration of a car be calculated?

Yes, the deceleration of a car can be calculated using the equation: a = (vf - vi)/t, where a is the deceleration, vf is the final velocity (usually 0 for a complete stop), vi is the initial velocity, and t is the time it takes for the car to come to a stop. This equation is based on Newton's second law of motion, which states that force is equal to mass times acceleration.

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