Which person exerted a force perpendicular to displacement?

AI Thread Summary
Jack and Jill are moving a boat, with Jack exerting a force of -80,0,290 N and Jill exerting 0,240,0 N. Jack does -3410 J of work, while Jill does no work (0 J). The discussion revolves around determining which person's force is perpendicular to the boat's displacement. The conclusion is that Jill's force vector is perpendicular to the displacement, as indicated by a zero dot product with the displacement vector. Therefore, Jill is the one who exerted a force perpendicular to the boat's movement.
Westin
Messages
87
Reaction score
0

Homework Statement



Jack and Jill are maneuvering a 3570kg boat near a dock. Initially the boat's position is [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png2,0,9[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngm and its speed is 4.6m[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngs. As the boat moves to position [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png12,0,0[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngm, Jack exerts a force of [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png-80,0,290[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngN, and Jill exerts a force [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png0,240,0[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngN.

How much work does Jack do?

WJack= -3410 J

How much work does Jill do?

WJill= 0 J

The second part of the question asks... Which person exerted a force perpendicular to the displacement of the boat?
a) Neither Jack nor Jill.
b) Both Jack and Jill.
c) Jill
d) Jack.

I was able to find the Work correctly but not I am just having a little trouble with the second part.
[/B]

Homework Equations



W=Fd[/B]

The Attempt at a Solution



At first, I thought it was Jack since Jack is the only person exerting Work. But then I thought Jill because she's exerting force, but just in the wrong direction (perpendicular?) so her work is 0. I only have one shot at this, is my reasoning ok for saying its Jill?[/B]
 
Last edited by a moderator:
Physics news on Phys.org
Hi there,

Firstly I took the difference between the starting and finishing vectors to get a movement vector of (10, 0, -9). Then there is the vector of force that Jill exerts (0, 240, 0).

Two vectors are perpendicular if their dot product is zero. So in this case the dot product is (10x0)+(0x240)+(-9x0) = 0+0+0 = 0.

So I'd say that yes, Jill's force vector is perpendicular to the displacement of the boat.

Steven Thomas.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top