Understanding Work and Energy: Solving a Physics Problem

In summary, the conversation is about a physics problem involving an object sliding on a flat surface with an initial speed of 30 m/s and coming to a stop in 10m. The coefficient of kinetic friction between the object and the surface needs to be determined. The individuals in the conversation discuss the equations for kinetic friction and kinetic energy, as well as the role of mass in the problem. The solution involves using the SUVAT equations for a body moving with constant acceleration and connecting the equations for kinetic friction and kinetic energy.
  • #1
kixtoby
8
0
Misplaced homework moved here by moderator, hence no template.
I have been stuck on this physics question:

An object is sliding along a flat surface with an initial speed of 30 m/s. What must the coefficient of kinetic friction be between the object and the surface if the object slides to a stop in 10m? What would be the speed of the object after it had only slide half that distance?

So I know that uk = Fk/FN
and that Fd = 1/2mv(initial)^2
but for the life of me I cannot figure out how to solve this problem.

Any help at all would be extremely appreciated.

I have done this:
F(10m)=1/2(9.8m/s2)(30m/s)
F(10m)=147m/s
F=14.7m/s

Am I at least on the right track?
 
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  • #2
kixtoby said:
and that Fd = 1/2mv(initial)^2

kixtoby said:
I have done this:
F(10m)=1/2(9.8m/s2)(30m/s)

Be careful in the substitution. Do it again. Also what are you trying to find by this?
 
  • #3
QuantumQuest said:
Be careful in the substitution. Do it again. Also what are you trying to find by this?
I am honestly not sure, I guess what I was trying to find was the force? But I need to find the coefficient of kinetic friction between the object and the floor. I guess I'm not sure what direction I need to go to solve this.
 
  • #4
kixtoby said:
I am honestly not sure, I guess what I was trying to find was the force? But I need to find the coefficient of kinetic friction between the object and the floor. I guess I'm not sure what direction I need to go to solve this.

Is the substitution you do correct? Please check it again
 
  • #5
QuantumQuest said:
Is the substitution you do correct? Please check it again
I noticed I forgot to square...was that my only error? I also was not sure about plugging in gravity, but no mass was given in the problem...
F(10)=1/2(9.8)(30)^2
F(10)=4410
F=441
 
  • #6
kixtoby said:
I noticed I forgot to square...was that my only error? I also was not sure about plugging in gravity, but no mass was given in the problem...

Squaring is second. First, you have substituted g for mass. This is wrong. Now, if m is not given, can you think of a way to not have it in your equation?
 
  • #7
QuantumQuest said:
Squaring is second. First, you have substituted g for mass. This is wrong. Now, if m is not given, can you think of a way to not have it in your equation?
Find a way to cross it out?
 
  • #8
kixtoby said:
Find a way to cross it out?

Cancelling is one way. What can you do and what can you find?
 
  • #9
What is the acceleration of the object? Have you drawn a free body diagram?
 
  • #10
Chestermiller said:
What is the acceleration of the object? Have you drawn a free body diagram?
how would I calculate acceleration from the information given? And yes, I did draw one, although I wasn't really sure what values to plug in...
 
  • #11
QuantumQuest said:
Cancelling is one way. What can you do and what can you find?
Can I multiply mass on each side to cancel them out? Then the equation would be: fd=1/2v^2
 
  • #12
kixtoby said:
Can I multiply mass on each side to cancel them out?

I don't understand this. Can you elaborate writing the equation?

EDIT: As a hint, what does the initial equation you wrote express in words?
 
  • #13
kixtoby said:
Can I multiply mass on each side to cancel them out? Then the equation would be: fd=1/2v^2
What is f? Previously you used F for force. If f is the same, the left hand side is energy, while right hand side is velocity-squared. That would be dimensionally impossible.
Relate FN to mass.
 
  • #14
QuantumQuest said:
I don't understand this. Can you elaborate writing the equation?
Original
haruspex said:
What is f? Previously you used F for force. If f is the same, the left hand side is energy, while right hand side is velocity-squared. That would be dimensionally impossible.
Relate FN to mass.
I am very lost. I don't think I understand this problem at all.
 
  • #15
kixtoby said:
Original

Original?

EDIT: Try to analyze the equation. What does it express? What can you substitute and what can you find from it in order to proceed to the solution of the problem?
 
  • #16
kixtoby said:
Original

I am very lost. I don't think I understand this problem at all.
What are the vertical forces? What is the vertical acceleration? If m is the mass, what equation does that give you?
 
  • #17
QuantumQuest said:
Original?

EDIT: Try to analyze the equation. What does it express? What can you substitute and what can you find from it in order to proceed to the solution of the problem?
Sorry, I didn't realize I had replied to you. I started to type and must have hit reply by mistake.
So for this problem, I know that friction does work: f*d which is is equal to the initial kinetic energy (1/2)mv^2
Then, F = mu * mg where my goal is to solve for mu (coefficient of kinetic friction).

I am still not understanding how to go about this without knowing the mass. Is there another hint you could give me? I almost think I am making this harder than it needs to be, but I just don't understand.
 
  • #18
Are you familiar with the SUVAT equations for a body moving with constant acceleration? If so, which one do you think you can you use in this problem to calculate the acceleration of the object (This part of the calculation does not involve Newton's laws. It is totally kinematic)?
 
  • #19
kixtoby said:
f*d which is is equal to the initial kinetic energy (1/2)mv^2
Then, F = mu * mg where my goal is to solve for mu (coefficient of kinetic friction
Do F and f stand for the same thing here? If so, why not connect those two equations, eliminating it? If not, what are they?
 
  • #20
kixtoby said:
I am still not understanding how to go about this without knowing the mass. Is there another hint you could give me?

Look at what haruspex asked in #19. You have to understand which force's work is f*d - as you write it, in your initial equation. Also, I recommend looking at it the way Chestermiller suggests in #18.
 

1. What is the difference between work and energy?

Work and energy are two related concepts in physics that are often confused with one another. Work is defined as the force applied on an object multiplied by the distance the object moves in the direction of the force. Energy, on the other hand, is the ability of a system to do work. In simpler terms, work is a measure of the effort put into a task, while energy is the capacity to perform work.

2. How can work and energy be calculated?

The formula for calculating work is W = F x d, where W is work, F is force, and d is distance. The unit for work is joules (J). Energy can be calculated using the formula E = m x g x h, where E is energy, m is mass, g is the acceleration due to gravity, and h is the height of the object. The unit for energy is also joules (J).

3. What is the relationship between work and kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is directly related to work, as work is required to change an object's motion and therefore its kinetic energy. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

4. How is potential energy related to work?

Potential energy is the energy an object possesses due to its position or configuration. It is also directly related to work, as work is required to change an object's position or configuration and therefore its potential energy. The work done against gravity is equal to the change in an object's potential energy.

5. What are some real-life examples of work and energy problems?

Some common examples of work and energy problems include lifting a heavy object, pushing a cart, riding a bike up a hill, and throwing a ball. In each of these situations, work and energy are being transferred and transformed to accomplish a task. Other examples can include using machines, such as a pulley system or a lever, to do work, or calculating the amount of energy needed to power a car or a household appliance.

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