# Work and Energy with 2 variables

• zenophysics
In summary, the car's velocity was calculated to be 2.89 m/s, which is equivalent to 6.46 mi/h. However, it is unlikely that the car would skid 88 m with that kind of velocity.
zenophysics

## Homework Statement

I feel as though I am missing information but I have been assured I'm not can anyone help me out?

At an accident scene on a level road, investigators measure a car's skid mark to be d=88m long. The accident occurred on a rainy day, and the coefficient of friction was estimated to be Mu=0.34. Use this data to determine the speed of the car, v when the driver slammed on (and locked) the breaks. Take g=10m/s^2.

## Homework Equations

What is the speed of the car?
F=ma Ft=m(delta)v

## The Attempt at a Solution

E1=E2+|W*Ffr|
mv^2/2=mv^2/2+|W*Ffr|

Last edited:
zenophysics said:

## The Attempt at a Solution

E1=E2+|W*Ffr|
mv^2/2=mv^2/2+|W*Ffr|

Initially the car has KE mv2/2

when the car has gone a distance of 88 m the velocity is zero.

So applying your second formula you will get

mv2/2 = Work done by friction.

You should know how to get the frictional force (so you will see why you don't need the mass of the car).

ok so if mv^2/v = WFfr and Ffr is the same as Fn*mu = mg*mu in the opposite direction then mass cancels out and leaves me with v^2/2=g*mu and when I solve for V I'm getting v=2.89ms. yes?

You're on the right track, but I see one error in your attempt to solve the problem.

Namely the statement in bold here:
E1 = E1'+W*Ffr

As a quick review, we know that

W = Fd

and

W = KE

Thus,

KE = Fd

Well, heat caused by friction is a form of kinetic energy, so we can surmise the following:

Eh = Ffrd

Can you see how this can apply to your equation?

Please note, however, that this only accounts for the transfer of your original kinetic energy into heat energy. That isn't a problem in introductory physics, but sound and other forms of energy may be important in later years.

zenophysics said:
ok so if mv^2/v = WFfr and Ffr is the same as Fn*mu = mg*mu in the opposite direction then mass cancels out and leaves me with v^2/2=g*mu and when I solve for V I'm getting v=2.89ms. yes?

2.89m/s is equivalent to 6.46 mi/h (10.4 km/h), it seems unlikely that a vehicle would skid 88 meters with that kind of velocity.

Where did you get the term " W*Ffr " and what does it define?

## 1. What is work and energy with 2 variables?

Work and energy with 2 variables refers to a physical system in which both work and energy are involved and are dependent on two different variables. This can include scenarios such as a moving object with changing speed, or a system with both kinetic and potential energy.

## 2. How are work and energy related?

Work and energy are closely related concepts in physics. In simple terms, work is the transfer of energy from one form to another, or the change in energy of a system. Therefore, work and energy have the same units and are interdependent on each other.

## 3. What are the units of work and energy?

The SI unit for both work and energy is joule (J). In some cases, the unit of work may also be expressed as Newton-meter (N⋅m) or kilogram-meter squared per second squared (kg⋅m²/s²).

## 4. How do you calculate work and energy with 2 variables?

The calculation of work and energy with 2 variables depends on the specific scenario and the variables involved. In general, work is calculated by multiplying force by displacement, while energy can be calculated using various equations such as the kinetic energy equation (KE = 1/2mv²) or the potential energy equation (PE = mgh). It is important to identify the variables and use the appropriate equations for the given scenario.

## 5. What are some real-life examples of work and energy with 2 variables?

There are many real-life examples of work and energy with 2 variables, such as a roller coaster moving at varying speeds (both kinetic and potential energy are involved), a car accelerating or decelerating (work is done by the engine to change the car's energy), or a person climbing a set of stairs (potential energy is converted to kinetic energy as they move upwards).

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