Estimating Speed Using Work and Energy Equations

  • Thread starter chaostheory1
  • Start date
In summary, the problem involves a race car with a mass of 1000 kg traveling at high speed on a wet concrete road. The driver locked the brakes and skidded 100 m, leaving visible marks on the road. The driver claimed not to have been exceeding 65 miles per hour. When trying to solve for the kinetic energy using the equation W=KE, the calculated workload for the brakes was over 300,000 N, which seemed impossible. However, this is because work and energy have different units (Joules vs Newtons). To solve for the speed, the work-energy theorem can be used by setting the work done by friction equal to the initial kinetic energy and solving for the speed.
  • #1
chaostheory1
3
0
I have honestly been trying to figure this problem out for over an hour now, and every time I try to solve for KE, I get a ridiculously high number.

Homework Statement


A race car having a mass of 1000 kg was traveling at high speed on a wet concrete road under foggy conditions. The tires on the vehicle later were measured to have µ = 0.55 on that road surface. Before colliding with the guardrail, the driver locked the brakes and skidded 100 m, leaving visible marks on the road. The driver claimed not to have been exceeding 65 miles per hour (29 m/s). Use the equation W=KE to estimate the driver's speed upon hitting the brakes.

Homework Equations


W=KE
KE=(mv^2)/2

The Attempt at a Solution


Every time I try to substitute and solve for KE, I'm getting a workload of over 300,000 N for the brakes, obviously impossible.
 
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  • #2
First of all, why is it "obviously impossible"? Secondly, how exactly are you getting that number? (i.e. show your work)
 
  • #3
chaostheory1 said:
Every time I try to substitute and solve for KE, I'm getting a workload of over 300,000 N for the brakes, obviously impossible.
Show exactly what you are doing.

One formula you didn't list is the one you need to compute the work done by friction. What would that be?
 
  • #4
Ok,
KE=(.5)(mv^2)
KE=(.5)(1000 kg * (29 m/s)^2)
KE=420,500 N
W=KE
How can I possible have 420,500 N of work done by brakes?
 
  • #5
Or am I doing this wrong...
Since the force that stopped the car was frictional; won't I have to calculate the frictional force using the Newton weight of the car? And use the frictional force in the KE and Work formulas?
 
  • #6
Work is not just equal to K.E.

The work-energy theorem says:

W = delta K.E. = [1/2 mv(final)^2] - [1/2 mv(initial)^2]
 
  • #7
chaostheory1 said:
Ok,
KE=(.5)(mv^2)
KE=(.5)(1000 kg * (29 m/s)^2)
KE=420,500 N
W=KE
How can I possible have 420,500 N of work done by brakes?
The unit of work/energy is Joules, not Newtons. And why do you think that that is an unreasonable amount of energy, or that brakes can't handle it? (Note that you don't need to calculate this energy to solve the problem.)

chaostheory1 said:
Or am I doing this wrong...
Since the force that stopped the car was frictional; won't I have to calculate the frictional force using the Newton weight of the car? And use the frictional force in the KE and Work formulas?
Absolutely. What does the friction force equal? (Symbolically, not numerically.) Set the work done by friction equal to the initial KE, then solve for the speed.
 

FAQ: Estimating Speed Using Work and Energy Equations

What is "W=FD with a µ Curveball"?

"W=FD with a µ Curveball" is a scientific concept that relates to the relationship between work, force, and distance, taking into account the effects of friction and air resistance on a spinning object, such as a baseball thrown with a curveball motion.

How is the µ Curveball related to "W=FD"?

The µ Curveball is a variable that represents the coefficient of friction on a spinning object, such as a baseball. When this variable is factored into the equation "W=FD", it allows for a more accurate calculation of the work done on the object.

Why is the µ Curveball important in scientific research?

The µ Curveball is important because it allows for a more precise understanding of the forces at play on a spinning object, which can be applied to various real-world situations, such as sports or aerodynamics. It also highlights the role of friction in these scenarios.

How does the µ Curveball affect the trajectory of a thrown object?

The µ Curveball can affect the trajectory of a thrown object by creating a force that acts perpendicular to the direction of motion, causing the object to curve. This is due to the uneven distribution of air resistance on the spinning object.

Can the µ Curveball be applied to other spinning objects besides a baseball?

Yes, the concept of the µ Curveball can be applied to any spinning object, such as a frisbee or a spinning top. The coefficient of friction may vary depending on the object, but the basic principle remains the same.

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