Phyisics help?!?

i have you at 10/10 so i liked you at one point, good luck in findiing help mN

i remember a bit of this from last year. it always help to draw a free body diagram of the forces on the object( at least what they taught me).

force of friction = (coefficient of friction) x (normal force)

sorry that is all i could remember. good luck!

first question

Work= Force x distance

since she is going a constant velocity, the only force you need to worry about is the friction.
the force of friction on a level surface would be the coefficient of kinetic friction x the normal force, and applies said force in the direction opposite of the motion.

The normal force is the same as the force that the girl puts on the ground, but opposite in direction, so 35kgx 9.8 m/s^2= 343 N, so the force of friction would be 343N x .20, or 68.6 N

So you know the force now, and you only have to multiply that by the distance to get the work, so 68.6N x 10m= 686 Nm

_____

For the second problem, I could have helped you in a second last semester, but this semester of physics has been electromagnetic fields, which are fucking ridiculous, and I have forgotten the formulas for an incline w/ friction, but it should be as simple as a free body diagram.

already had you 10/10 karma...wish it could go higher.  thanks!

no problem, what better way to procrastinate from studying my calc 2 exam than sitting on NS and doing other people's math instead?

ok i havent done this in almost 2 years so bear with me. i feel like we need to use tension here but i might be wrong. i feel like im missing something, so dont trust me 100% anyways,

Ff = mu ( normal force of weight)
Ff = 0.20(35 x 9.81)
Ff = 68.67 (make -ve since it opposes motion)

work = force (dot) distance = (F)(d)cos angle
work = (-68.67)(10)cos(180) = 686.7 joules of work




Ff = mu(weight)cos 20
= 0.275((3 x 9.81)cos20)
Ff = (-ve) 7.61 N

work = Fdcos180
w = (-7.61)(1.5)cos 180
work = 11.415 joules

+++K

1st problem's all good

But for 2nd problem :

Ff =-7.6 N      Thats cool but his weight is going down the hill with a force that would be equal to +10 N if there was no friction ->Fw= 3kg*9,81*sin(20)=10N

so global force applied on this block is f=10N-7,6N=2,4N

And work should be : W=2,4N*1,5m=2,6J

If you want to know the work done by friction then it should be w = (7.6)(1.5)cos 180=-11,4J (The - sign mean the work's done in a direction opposed to your references axes (we suppose the X axe is growing in the direction the weight's moving)

and if you want the work done by the weight it should be w = 10N*1,5*cos(0)=15J ... 15J-11,4J=2,6J

Im not rly shure tho.. im confused with this old definition of work.. but w-e the overall energy lost by the block will be 11,4J, the overall potential energie lost by the block is 15J. The 2,6J of energy left at the end is cinetic energie.. the potential energy was either lost due to friction or transformed in cinetic energie (mouvement).

Later knowing your block gained 2,6J or cinetic energie you can determin his speed at the bottom of that 1,5m hill will be 1,32m/s with

Ec=(m*v^2)/2

2,6J=(3*v^2)/2 -> v=(5,2/3)^1/2=sqrt(5,2/3)=1,32 m/s

Ya come back next year for electromagnetic fields and 2D collisions

if anyone wants to help with this one i would appreciate it:


A series RLC circuit with 48 , 240 , and 4.0  is connected across a sine-wave generator whose peak output voltage is independent of frequency.

Find the minimum frequency of the frequency range over which the peak current will exceed half its value at resonance. Hint: You can solve this problem graphically or, with appropriate algebraic manipulations, using quadratic equations.

hey jeffrey if you be needing the physics help
just give me a call