Shown below is a graph of the 1-dimensional, net force as a function of position

Question

Shown below is a graph of the 1-dimensional, net force as a function of position, F(x), that you exert on a particle of mass 14 kg.

Note 1: It may be easier to solve these problems in a different order than they are asked.
Note 2: F = 0 at x = 6.087 m.

The particle starts at x = 1 m with a velocity of 6.570 m/s and reaches x = 20.00 m under the action of the force in the graph.

What is the impulse that you give to the particle as it moves from x = 1.00 m until it reaches x = 20.00 m. Remember that impulse is a vector, so in 1-dimension it can be either positive or negative.

For the situation described above, what is the work done on the particle by the force from x = 1.00 m to x = 20.00 m.

For the situation described above, what is the velocity of the particle when it reaches x = 20.00 m.

For the situation described above, what is the impulse that the particle gives to you as it moves from x = 1.00 m until it reaches x = 20.00 m. Remember that impulse is a vector, so in 1-dimension it can be either positive or negative.

Explanation / Answer

lets first find out the expression for force from x=1 to x=20 m:

part 1: from x=1 m to x=14 m

at x=14 m , F=-14 N

at x=6.087 m, F=0 N

as it is a striaght line, the equation for force vs position will be:

(F-0)/(x-6.087)=(0-(-14))/(6.087-14)

==>F/(x-6.087)=-1.7692

==>F=-1.7692*x+10.769

part 2: from x=14 m to x=20 m: Force F=-14 N

then total work done on the particle=integration of F.dx

as F is a piecewise function, lets integrate for each part and then add to get the total work done

work done during part 1:

F=-1.7692*x+10.769 N

==>F.dx=-1.7692*x*dx+10.769*dx

==>integration of F.dx=-1.7692*0.5*x^2+10.769*x

applying limits from x=1 to x=14 m

we get integration of F.dx=-32.5 J

hence work done during part 1=-32.5 J

part 2:

as force=-14 N, total work done =F*(20-14)=-14*6=-84 J

hence total work done=-32.5-84=-116.5 J

as there are no external forces,

change in kinetic eenrgy=total work done on the particle

if final veloicty at x=20 m is v1,

then 0.5*mass*(final veloicty^2-initial velocity^2)=total work done

==>0.5*14*(v1^2-6.57^2)=-116.5

==>v1^2-6.57^2=-16.643

==>v1^2=26.522

==>v1=5.15 m/s or v1=-5.15 m/s

[we have two solutions from which only one is correct. to understand which one, we have to check whether veloicty has become zero at any point of time

as we know, force=m*v*dv/dx

where m=mass of the particle=14 kg

for part 1,

14*v*dv/dx=-1.7692*x+10.769

==>14*v*dv=-1.7692*x*dx+10.769*dx

==>14*v^2/2=-1.7692*0.5*x^2+10.769*x

setting initial limit as x=1 and v=6.57 m/s and final limits as x=d and v=0 (so that we can find at which x=d, v=0)

-6.57^2/2=-1.7692*0.5*(d^2-1)+10.769*(d-1)

==>-21.582=-0.8846*d^2+0.8846+10.769*d-10.769

==>0.8856*d^2-10.769*d-11.698=0

solving for d we get d=13.16 m

hence indeed the veloicty has become 0 and as force continues to be negative, veloicty will also become negative.]

hence the proper answer is v1=-5.15 m/s

answers:

question 1:

impulse provided=change in momentum

=mass*(final speed-initial speed)

=14*(-5.15-6.57)=-164.08 kg.m/s

question 2:

total work done on the particle is -116.5 J.

question 3:

velocity of the particle when it reaches zero is given by -5.15 m/s

question 4:

repeat of question 1

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