Racecar accelerating against the forces of friction....

[Baran]

Homework Statement

750kg racecar, starting from rest, experiences an applied force of 9800N. Frictional forces both wind resistance and rolling friction, push back against the car with 2100N. If the car is racing a 402m, what is the final speed of the car? Answer using energy and work laws, not kinematics.

Relevant Equations

No equations given

I don't know what to do.

 

[Drakkith]

What are your energy and work laws?

 

[Baran]

What are your energy and work laws?

I guess that would be the formulas,

W=F(Total Distance)
Ek=1/2(m)(v)^2
Wnet=[(m)(vf)^2]/2 - [(m)(vi)^2]/2
Em intial=Em final
Eg=mghThat's everything that I could think of.

 

[etotheipi]

Em intial=Em final

The mechanical energy of a system is only conserved in the absence of external work (plus the absence of internal dissipative forces ). If your system is just the car, we have external work and this equation won't apply.

This problem can be attacked by the standard work energy theorem, which you have correctly stated as $W = \Delta T = \frac{1}{2}m{v_f}^2 - \frac{1}{2}m{v_i}^2$. But first you've got to figure out the total work done by all forces acting on the car.

Can you figure out what the total work done is here? Be careful about signs, if the force is in the opposite direction to the displacement the work it does will be negative...

 

[Baran]

The mechanical energy of a system is only conserved in the absence of external work (plus the absence of internal dissipative forces ). If your system is just the car, we have external work and this equation won't apply.

This problem can be attacked by the standard work energy theorem, which you have correctly stated as $W = \Delta T = \frac{1}{2}m{v_f}^2 - \frac{1}{2}m{v_i}^2$. But first you've got to figure out the total work done by all forces acting on the car.

Can you figure out what the total work done is here? Be careful about signs, if the force is in the opposite direction to the displacement the work it does will be negative...

Would the total work be +7 700N?

 

[etotheipi]

Would the total work be +7 700N?

Remember that work has dimensions of energy, and $7700\text{N}$ has dimensions of force. That said, $7700\text{N}$ is the resultant force on the car. What do you need to multiply this by to get the work done by the resultant force?

 

[Baran]

The total distance, so
W=402mx7700N
=3.1x10^6 J

 

[haruspex]

The total distance, so
W=402mx7700N
=3.1x10^6 J

Yes. So can you find the speed?

 

[Baran]

Yes. So can you find the speed?

I'm not sure how to

 

[Drakkith]

I'm not sure how to

Is there a formula that links the speed/velocity of an object to its energy?

 

[Baran]

Is there a formula that links the speed/velocity of an object to its energy?

I think it is
Ek=1/2(m)(v)^2

so would I sub in the values like this?

3.1x10^6 J=1/2(750kg)(v)^2

then rearranged I think it would be

v=2Ek/m (the right hand side should have a square root, but I don't know how to make that symbol)Also sorry about the late reply, I was busy with some other IRL things, and forgot about this :P.

EDIT:

I'm starting to think that it may be Wnet=[(m)(vf)^2]/2 - [(m)(vi)^2]/2 and then I would have to rearrange for vf, but this seems very complex to rearrange.

 

[jbriggs444]

I think it is
Ek=1/2(m)(v)^2

so would I sub in the values like this?

3.1x10^6 J=1/2(750kg)(v)^2

Instead of substituting in and then solving, try the other way around -- solve then substitute. You'll end up with the same result either way, but algebra is easier with variable names than with numeric values. And errors are easier to find and correct.

Typesetting instructions can be found here.

Let's type set that starting equation using TeX. Putting a double dollar sign in front and a double dollar sign behind you get to render a mathematical equation centered on the line.

This ($$E_k=\frac 1 2 mv^2$$) becomes this:$$E_k = \frac 1 2 mv^2$$One can also do it with doubled hash marks to get inline equations. This (##E_k=\frac 1 2 mv^2##) becomes $E_k=\frac 1 2 mv^2$

Let's solve the above equation for $v$. Multiply both sides by 2 and divide both sides by m. That gives us $$\frac{2 E_k}{m} = v^2$$ Now take the square root of both sides and swap right and left yielding $$v = \sqrt{ \frac{2 E_k}{m} }$$ But of course, you already knew that.

Now you can sutstitute in 3.1 x 10⁶ J for $E_k$ and 750 kg for m yielding $$v=\sqrt{\frac{2 \times 3.1 \times 10^6 \text{J}} {750 \text{kg}}}$$

[Cheating to display back to back dollar signs and hash marks is tricky. I pulled down the little paint droplet icon to pick a color for the second dollar sign, making it explicitly black. Enclosed in a color tag that way, the character was no longer able to activate the equation rendering engine]

 

[Baran]

Instead of substituting in and then solving, try the other way around -- solve then substitute. You'll end up with the same result either way, but algebra is easier with variable names than with numeric values. And errors are easier to find and correct.

Typesetting instructions can be found here.

Let's type set that starting equation using TeX. Putting a double dollar sign in front and a double dollar sign behind you get to render a mathematical equation centered on the line.

This ($$E_k=\frac 1 2 mv^2$$) becomes this:$$E_k = \frac 1 2 mv^2$$One can also do it with doubled hash marks to get inline equations. This (##E_k=\frac 1 2 mv^2##) becomes $E_k=\frac 1 2 mv^2$

Let's solve the above equation for $v$. Multiply both sides by 2 and divide both sides by m. That gives us $$\frac{2 E_k}{m} = v^2$$ Now take the square root of both sides and swap right and left yielding $$v = \sqrt{ \frac{2 E_k}{m} }$$ But of course, you already knew that.

Now you can sutstitute in 3.1 x 10⁶ J for $E_k$ and 750 kg for m yielding $$v=\sqrt{\frac{2 \times 3.1 \times 10^6 \text{J}} {750 \text{kg}}}$$

[Cheating to display back to back dollar signs and hash marks is tricky. I pulled down the little paint droplet icon to pick a color for the second dollar sign, making it explicitly black. Enclosed in a color tag that way, the character was no longer able to activate the equation rendering engine]

Thank you for your help, as well as all the other people who helped me solve this question. I really do appreciate all the help that I've received!

 

[gmax137]

Thank you for your help, as well as all the other people who helped me solve this question. I really do appreciate all the help that I've received!

So, @Baran, do you know what the original question meant when it said, "Answer using energy and work laws, not kinematics"?

What would the approach be if it had asked the opposite ("solve using kinematics")?

 

[jbriggs444]

Do you know what the original question meant when it said, "Answer using energy and work laws, not kinematics"?

What would the approach be if it had asked the opposite ("solve using kinematics")?

Edited to remove spoiler. Oops.

 

[gmax137]

I was kind of hoping the OP might answer.