Given F=4i-2k; P(0,1,0) and Q(4,0,1), find the work W done by the force F in moving a particle in a straight?
547865
I guess that the above answer has a typo\nPQ is Q-P = (4,0,1)-(0,1,0)= (4,-1,1)\nW = F . PQ = (4,0,-2).(4,-1,1) = 16-2= 14 units
398633
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
99566
p=q*d\nq=1.60*10^-19 coulombs\nd=4.30*10^-3 m\nso\np = q*d = 1.60*10^-19 * 4.30*10^-3 = 6.89*10^-22 coulomb*meters
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
884829
The work done by friction is the difference between the kinetic energy of the skier at the bottom of the slope and the potential energy of the skier when he stops.  So KE (initial) = PE (final) + W.  KE is 0.5*m*v^2 and PE is mgh, where h is the height at rest.  W is Fd = umgd*cos(theta), where u is the coefficient of friction, d is the path length traveled, and theta is the angle of inclination.  The height h will be d*sin(theta).  The equation becomes 0.5mv^2 = mgd*sin(theta) + umgd*cos(theta).  The  mass is constant and unknown, so we can factor it out, leaving 0.5v^2 = gd*sin(theta) + ugd*cos(theta).  u is your only unknown, so you can solve algebraically.
900660
You must compute the slope at which the particle is moving at\nthat point.  Then the ratio of the absolute speed to the horizontal \nspeed is equal to the ratio of the hypotenuse of the slope to adjacent side.  And of course the slope is computed by the derivative's value at x=4.  So\n  y=x^1/2\n  dy/dx = 1/2(x^(-1/2)= 1/(2x^1/2)   \n  dy/dx=1/4    at x=4  So  the ratio of the  vertical speed to the horizontal speed is 1/4 and the proportional hypotenuse  in the direction of the particle(i.,e. the curve) is\n (1^2 + 4^2)^1/2 =  17^1/2\n\n   So    (17^(1/2))/4 = v/3cm/s   where v is the velocity in direction\nof the curve is\n\n       v=(3/4)(17^(1/2))\n\nStuart's approach is slightly different and fascinating to me\nbut also right.\n\n  Garipopki is also right but a bit work intensive I would say.\n\n Garipopki's answer is closer to mine and I'm backing off\nfrom supporting stuart. Work intensive or not, correctess\nsupercedes all.\n\n\nAt this point I'm responding to garypopki's comment that I'm\ncomputing the velocity along the curve and not the velocity\naway from thr origin. Good point. So the velocity away from\nthe origin is cos of the angle between the two directions times\nthe absolute velocity of the particle which is in the direction of the curve.\n    Angle of the curve at the point (4,2) is  α   where\n  tanα = 1/4,  so α=14d (degrees, this is physics so I won't use\nradians)\n\n   and the angle from the origin from which the particle is receding\n is  β where tanβ = 1/2,  so β=26d   \n\n   26d-14d=12d \n\n   So the component of velocity away from the origin is the absolute velocity times the cos of the angle between them.\n\n    Absolute velocity is  3.0923 so velocity from the origin\nis cos(12)3.0923 = \n    3.024cm/s  \n\nwhich I have to admit is\nawfully close to the 3.018 that you'd get from garypopki's\n(27/20)5^(1/2)
734494
(a) Work done by gravity = change in potential energy of the book. Now change in potential energy = m*g*h2 - m*g*h1=2*10*(10-1.5) =  170J. So, work done by gravity is 170J\n(b)Change in gravitational potential energy = work done = 170J as explained above.\n(c) Potential energy when textbook is released = m*g*h1 = 2*10*10=200J\n(d) Potential energy when it reaches the hands = m*g*h2 = 2*10*1.5=30J\n(e) If U =100J at ground level, work done by gravity = 170J, the work done does not change with change in reference.\n(f) change in potential energy also remains 170J\n(g) U at release point = 100 + 2*10*10 = 300J\n(h) U at hands = 100 + 2*10*1.5 = 130J
989843
net force IN THE DIRECTION OF DISPLACEMENTis (1.7*10^6 * cos 14)*2 \n= 3.3* 10 ^ 6 N\nwork done = 3.3*10^6*860=2.838 * 10^9 Joules
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
546224
Energy = force x distance\n=>  Energy = (mass x acceleration) x Distance\n=>  Energy = mass x acceleration x Distance\n=>  Energy = mass x (speed / time) x Distance\n=>  Energy = mass x speed x (Distance/time)\n=>  Energy = mass x speed x speed\n=>  Energy = mass x (speed)²
862116
KE = 0.5*m*(v)^2 = 0.5*0.023*(848)^2 = 8269.696 J\n\nThis must = work to accelerate the bullet\n\nW = F*d = 8269.696\n\nF= 8269.696/0.803 = 10298.5 N
955845
The object has kinetic energy (1/2)*m*v^2 when first at x=0. The work friction does consumes all that energy as the spring compresses and pushes the object back to x=0. Let the distance the spring is compressed be x1. Then \n(1/2)*m*v^2 = Ff*2*x1 = mu*m*g*2*x1\n\nWhen the spring is compressed to x1, the friction has decreased the energy in the system by half and the remaining half is all stored in the spring compression because the object has come to a stop - about to start the rebound. So\nFspring*x1 = k*x1^2 = (1/4)m*v^2\nIf you solve the 1st equation for x1 and substitute the result into the 2nd, you should get the required result.
63995
ok, all you have to do is figure out how much work it should have been doing, then compare it to the amount of work it actually did by using a percentage. so first of all, W=Fx(work equals Force times distance) that means that W=(12)(.3)=3.6 J of work.  Ok, so since it only did 3.2 J, that means .4J were lost somewhere.  the efficiency is calculating the efficiency in terms of what percentage of the work that was actually applied was translated into the work energy that was done on the lid.  the way you do this is just divide the amount of work it did by the amount of work it should have done(in an ideal solution) and then multiply by 100 in order to find the percentage of work it actually did in relation to the work it should have done.  3.2/3.6= .88889(x 100) = 88.89%.  \n\nso basically, in cases with efficiency, you divide the number you solved for by the number they gave you. a good way to figure out which you divide by which is just to randomly try dividing and if the answer is above 1(100%), that's impossible so you switch them ; )
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
861187
0.4*(100kg*g+F*sin(x)) = F*cos(x). ignoring any rotational component