Work...1

[nycmathguy]

Section 6.4

Can you please do 70 as a guide for me to do 69?

 

[nycmathguy]

Must I use the following to answer 69 and 70?

W = work done

W = (magnitude of Force)( distance from P to Q)

 

[MathLover1]

70.

P(1,3), Q(-3,5), v=-2i+3j

W=||v||*PQ

||v||=sqrt((-2)^2+3^2)=sqrt(13)
PQ=sqrt((-4)^2+2^2)=2sqrt(5)

W=sqrt(13)*2sqrt(5)
W=16.13

 

[nycmathguy]

70.

P(1,3), Q(-3,5), v=-2i+3j

W=||v||*PQ

||v||=sqrt((-2)^2+3^2)=sqrt(13)
PQ=sqrt((-4)^2+2^2)=2sqrt(5)

W=sqrt(13)*2sqrt(5)
W=16.13

OMG! It's so easy. I will do 69 later today. We move on to Section 6.5.

 

[Country Boy]

It is, unfortunately, wrong. Work is "magnitude of force vector times distance'' only if the motion is in the same direction as the force vector. Going from (1, 3) to (-3, 5) is moving in the direction of the vector <-3- 1, 5- 3>= <-4, 2>, which is not in the same direction as the force vector, <-2, 3>.

One interpretation of this is that there is some additional "restraining force" preventing the object from going in the direction of the force. For example, the object might be bead on a wire that can only move along the wire. Whatever force is applied to the bead, only the component of the force parallel to the wire can move the bead. The component perpendicular to the wire is cancelled by the "retraining force" of the wire itself.

The correct value of the work is given by the dot product of the motion vector and the force vector, <-4, 2> .< -2, 3> = 8+ 6= 14.

 

[nycmathguy]

70.

P(1,3), Q(-3,5), v=-2i+3j

W=||v||*PQ

||v||=sqrt((-2)^2+3^2)=sqrt(13)
PQ=sqrt((-4)^2+2^2)=2sqrt(5)

W=sqrt(13)*2sqrt(5)
W=16.13

Country Boy said that your answer is wrong. You say?

 

[nycmathguy]

It is, unfortunately, wrong. Work is "magnitude of force vector times distance'' only if the motion is in the same direction as the force vector. Going from (1, 3) to (-3, 5) is moving in the direction of the vector <-3- 1, 5- 3>= <-4, 2>, which is not in the same direction as the force vector, <-2, 3>.

One interpretation of this is that there is some additional "restraining force" preventing the object from going in the direction of the force. For example, the object might be bead on a wire that can only move along the wire. Whatever force is applied to the bead, only the component of the force parallel to the wire can move the bead. The component perpendicular to the wire is cancelled by the "retraining force" of the wire itself.

The correct value of the work is given by the dot product of the motion vector and the force vector, <-4, 2> .< -2, 3> = 8+ 6= 14.

I think you should let Mathlover1 know that she is wrong. Are you sure? She is an excellent mathematician.

 

[MathLover1]

Country Boy was right, I just made mistake

it should be:

W=F*PQ

v= <-2i,3j> => F=<-2,3>

displacement in this case is given by PQ:

PQ=q−p=<−3,5> - <1,3>= <-4,2>

W=F*PQ

W=<-2,3>*<-4,2>

W=-2(-4)+3*2

W=8+6

W=14J ( joules)

 

[nycmathguy]

Country Boy was right, I just made mistake

it should be:

W=F*PQ

v= <-2i,3j> => F=<-2,3>

displacement in this case is given by PQ:

PQ=q−p=<−3,5> - <1,3>= <-4,2>

W=F*PQ

W=<-2,3>*<-4,2>

W=-2(-4)+3*2

W=8+6

W=14J ( joules)

It's ok. We all make mistakes.