Work Problem/Graph Given/Find final velocity.

In summary, the graph shows the net external force component F cos θ along the displacement as a function of the magnitude of the displacement s for a 65 kg ice skater. The work done by the net force component on the skater is 93 Joules from 0 to 3.0 m and 0 Joules from 3.0 m to 6.0 m. To find the speed of the skater at s = 6.0 m, you can use the equation W = ΔKE and solve for the initial KE, then use the equation KE = 0.5mv^2 to find the final velocity.
  • #1
myxomatosii
80
0

Homework Statement



The graph shows the net external force component F cos θ along the displacement as a function of the magnitude of the displacement s. The graph applies to a 65 kg ice skater.

http://img9.imageshack.us/img9/3441/0671.gif (a) How much work does the net force component do on the skater from 0 to 3.0 m?
93 Joules (Correct)

(b) How much work does the net force component do on the skater from 3.0 m to 6.0 m?
0 Joules (Correct)


(c) If the initial speed of the skater is 2.4 m/s when s = 0, what is the speed when s = 6.0 m?

Homework Equations



W=F(Δx)
KE=.5mv2

The Attempt at a Solution



For part C.

Here is my attempt.

Firstly, I do not understand the reason that FCosΘ was used as a Force instead of simply F, due to the way the graph is drawn, I chose to simply represent it as F, which may have been my mistake, but I doubt that.

I know the WorkTotal done on the skater was 93 Joules. So I thought that perhaps the change in Kinetic Energy would be equivalent to 93 Joules, allowing me to solve for Vf.

So I tried that below.

FCosΘ=31N

W=FCosΘ(Δx)

FCosΘ(Δx)=.5mvf2-.5mvi2

.5mvf2=FCosΘ(Δx)+5mvi2

vf2=(2(FCosΘ(Δx))/m) + vi2

which means to solve for vf I would use.

vf=((2(FCosΘ(Δx))/m)0.5)+vi

Yet that answer is wrong. I think my concept of this problem is wrong..

For Δx I used 3 instead of 6, since the force drops off to zero and the environment is frictionless, so it shouldn't make a difference.
 
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  • #2
Your last step is wrong.
Check it.
 
  • #3
Seems a little complicated.

W = ΔKE

What is the initial KE?

Just add the Joules from a) and then solve for velocity?
 

1. What is a work problem?

A work problem is a type of physics problem that involves calculating the amount of work done on an object. This is usually represented by the equation W = Fd, where W is work, F is force, and d is displacement.

2. How do you solve a work problem?

To solve a work problem, you need to know the values of force and displacement, and plug them into the equation W = Fd. Make sure to use consistent units for force and displacement, such as Newtons and meters.

3. What is a graph given in a work problem?

In a work problem, a graph is often given to represent the relationship between force and displacement. This graph typically shows a straight line with force on the y-axis and displacement on the x-axis. The slope of the line represents the force and the area under the line represents the work done.

4. How do you find the final velocity in a work problem?

To find the final velocity in a work problem, you can use the equation vf = vi + (2W/m), where vf is the final velocity, vi is the initial velocity, W is the work done, and m is the mass of the object. Alternatively, you can use the equation vf = √(v^2i + 2ad), where a is acceleration and d is displacement.

5. Can you use a work-energy theorem to solve a work problem?

Yes, the work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. This can be useful in solving work problems, as it allows you to relate the work done to the final velocity of the object.

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