What's the Work and Speed of a Particle at Different Positions?

[eventob]

Homework Statement

The graph shows the acceleration of a particle with mass 2.00kg, as a function of its position x. The particle experiences an external force Fxi. It starts from rest at x=x0=0m.

How much work is done on the particle at:
x=4.00m
x=7.00m
x=9.00m

What is the particles speed, and direction, in the same three positions?

This is not the original problem statement, but I had to translate it in order for you guys to understand it I guess. :)

Homework Equations

Not sure. I can't use the kinemtic equations, since my acceleration is not constant, but a function of the particles possition. Also, I don't have the information needed to use the work-energy-theorem: change in kinetic energy=work done=(1/2)m*v^2.

The Attempt at a Solution

Really not sure where to start. I'm used to having a graph of the force vs. position, and then I can just calculate the area under the f_x(x) curve, but I can't do that here. Is there an easy way to go from acceleration to force? Was thinking about Newtons second law, because then the units would work out.

Thanks for your time. :)

 

[issacnewton]

hi

you have a graph of a Vs x. first what you can do is write the equation of a(x) for different
parts of the graph. for example, from x=0 to x=1 we have a=6x (use coordinate geometry to
get the equation of the line). from x=1 to x=4 we have a=6 and so on. from Newton's second law we see Net force = ma so you can get the force acting on the right direction.
for example from x=0 to x=1, F=ma = 6mx. so

\vec{F}=6mx\, \hat{i}
and work done would be
W=\int \vec{F} \cdot \vec{dx}
W=\int F\hat{i} \cdot dx \hat{i}
W=\int^1_0 6mx\,dx

and so it turns out from x=0 to x=1 that \inline{W=6 J}...you can work out in the same way for the other parts. the net work done from x=0 to x=4 would be the work done from x=0 to x=1 plus the work done from x=1 to x=4...

 

[eventob]

Thanks a lot. :)

 

[issacnewton]

for getting velocity, you need to express force in terms of displacement...
F=ma = m\frac{dv}{dt}

F=m\frac{dv}{dx}\frac{dx}{dt}

F=m\frac{dv}{dx}\, v

F=mv\frac{dv}{dx}

 

[eventob]

Just to make sure I got it right:
a_x(x) = 6x, x in[0,1]
a_x(x) = 6, x in[1,4]
a_x(x) = -6x, x in[4,6]
a_x(x) = -6, x in[6,8]
a_x(x) = 6x, x in[8,9]

Then I just integrate each of these portions of the function with respect to x, and add them together to get the total work done on the object? I'm just asking, because intuitively I thought that the work done from x=4 to x=5 would be the same as from x=0 to x=1, just by looking at the graph, since the lines have the same magnitude and opposite directions.

When i integrate the part x in[4,5] I get = int(Fxi*dxi)=int(Fx dx) with the lower limit x=4 and upper limit x=5 which gives me -12N*(1/2)(5^2-4^2) = -54J

:)

 

[issacnewton]

its good to have intuition but math is far more reliable... anyway your equations
no 3 and 5 are wrong... work them out...use the endpoints of these line segments to
find the slope and then the equation of the line..

 

[eventob]

Slope=delta y / delta x

= -6-(-6) / (6-4) = -6= m

y-6=m(x-4)
y=-6x-6*(-4)+6
=-6x+30
a(x)=y(x)

Thanks again. I really appreciate it.