What is the work done by the horizontal force on the object?
359330
shouldn't you subtract the work done by gravity\nfirst?\n\nwork By Horizontal Force = 55Nm - work By Gravity
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
913519
The gravitational force or the weight of the block is 13x 9.8 = 127.4 newton.\n\nWork done by the rocket against this gravitational force is “Force x displacement”\n\n“Force x displacement” = 127.4 x 11m = 1401.4 joules.\n\nWork is done against gravity and hence the gravity has lost some energy and is equal to 1401.4 joules.\n\n\nSince the block is moving with constant speed or velocity the net force acting on the block is zero.  Therefore the net work done by all the forces is zero.\n\n\nFor the part d:\n\nThe net force acting on the block is [130.68 -127.4] = 3.28newton.\n\nThis force of 3.28 newton does a work of moving the block through 11m.\n\nHence the work done on the block is 3.28 x 11 = 36.08 joule.\n\nThis work imparts kinetic energy to the block and hence the kinetic energy of the block is 36.08 joule.\n\nThe acceleration on the block is force/ mass = [3.28/ 13]\n\nInitial speed is zero; acceleration is [3.28/ 13] m/ s^2; distance traveled is 11m.\n\nThe final speed is found from v v = uu  + 2a s = 2as since u= 0.\n\nv v = 2 x [3.28/ 13] x 11\n\nThe kinetic energy of the block is 0.5 m v v = 0.5 x 13 x 2 x [3.28/ 13] x 11\n = 36.08 joule
509011
WORK:\nwork is said to be done when force is applied on a body and the body covers some distance in the direction of the applied force.\n\n                          W=F*S,WHERE:\nF=force applied\nS=distance\nNOW EXAMINING UR QUES\nif u lift an object in the direction of the force applied ,then yes work is done and if u lift it 3 times along the direction of applied force then work is tripled.\n\nONE SITUATION CAN BE THERE:\nif work done when force is not along the direction of motion-\nlet the angle of work be cos(theta)\n        cos(theta)0 =1\n        cos(  "  )30=(root)3/9\n        cos(  "  )40=1/(root)2\n        cos(  "  )60=1/2\n        cos(  "  )90=0
905836
So basically we're resolving all forces (no friction forces) for their horizontal components, summing them and then calling this the resultant force Fww that the wall exerts on the wedge.\n\nFirst step: resolve the forces acting on the block (of mass M) on the plane at angle theta to the horizontal\n\nF1 = normal force to the plane = Mgcos(theta)\nF2 = force in direction of (tangent to) the plane = Mgsin(theta)\n\nSecond step: resolve both forces found above into their horizontal components (i.e. normal to the wall)\n\nF1x = Mg sin(theta) cos(theta)\nF2x =Mg cos (theta) sin (theta)\n\nFww = F1x + F2x\nFww = 2Mg cos(theta) sin (theta)
880761
People above neglected the vertical component of 460 N.  This must be added to the weight of the box.  Also, not sure why they took 350 N and multiplied it by acceleration.\n\nso F_vert=460sin35.6=267.78 N (downward)\n     F_hor=460cos35.6=374.0 N\n\ncalculating F_frict=mu x (F_normal)\n\nF_normal=weight of box + F_vert = 267.78+350=617.78\n\nso F_frict=0.57x617.78=352.13 N\n\nnow the net force on the box = 374.0 - 352.13 = 21.87 N\n\nand we need mass of the box, m=350/9.81=35.7 kg\n\nso a=21.87/35.7=0.6126 m/s2\n\nfinally,  d=V_initial x t + 0.5 x a x t^2, V_initial=0\n\nso t=((4.2)/(0.5x0.6126))^0.5=3.7 sec\n\n**********************************************************\n\nHowever, should you wish to ignore the vert component\n\nF_frict=mu x (F_normal)=0.57 x 350 = 199.5 N\n\nso net horizontal force is:\n\nF= 374.0 - 199.5 = 174.5 N, a=174.5/35.7= 4.89 m/s2\n\nplug a into:\n\nd=V_initial x t + 0.5 x a x t^2, and solve for t\n\nt=((4.2)/(0.5x4.89))^0.5 = 1.31 sec
200510
The answer given by "Amanda" is actually quite funny, I like it.\n\nWork = Force * Distance applied\nIf a constant force of 45 Newtons is applies over a distance of 4 meters, the work done would be 180 Joules.\nI hope we can assume this 45 Newton force is being applied parallel to the ramp due to the weight (force of gravity) of the object.
354034
Forget the 6000N horizontal for for a second...\n\nYou have 5000N acting downward (negative) and 7000N acting upward (positive). The total net force acting vertically to the A/C is 7000-5000= +2000N so upward. We know that the horizontal force of 6000N is acting at a 90 degree angle to the upward force so the resulting forces make a right triangle. Use Pathagorean theorem.\n\n2000^2 + 6000^2 = x^2\n\nx is your MAGNITUDE.\n\nNow find the angle:\n\nUse trigonometry: \n\nTan(Angle) = opposite/adjacent\nwe know the opposite is 2000 and the adjacent is 6000 so,\ninverse Tan(2000/6000) = 18.4 degrees, upward and forward.
1179191
Work done by the force 2.34 N is, ‘force x distance through which the force is moved’.\n\nTherefore, it is = 2.34 N x 1.41m = 3.2994 joule.\n\nWork done by the frictional force 0.55 N is, ‘force x distance through which the force is moved’.\n\nTherefore, it is = 0.55 N x 1.41m = 0.7755 joule.\n\nThe last question is vague. \n\nThe total work is the work done by the force 2.34 N and it is equal to 3.2994 joule. \n\nOut of this total work a part (3.2994 -0.7755 = 2.5239) joule is used to accelerate the book and 0.7755 joule is dissipated as heat energy.
748800
1)c because the forces are equal (a also for the same reason 0 is constant)\n2)b because the box has an unbalanced force in the horizontal direction and to the right\n3)d because the unbalanced force is to the right\n4)d because there is more force to the right\n5)b there is an unblanced force to the right
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
859502
What you are likely looking is W=F*D \n\nWork = A Force * A Distance\n\nCheck out wikipedia for additional help:\nhttp://en.wikipedia.org/wiki/Mechanical_work
234860
See the fig in the web site. Instead of lifting the load through a vertical height 'h' we are moving the load through a longer distace 'l' along the plane.\nIn any machine work will be the same.\nWork is the product of force and distance.\nIf the load is moved vertically we would have done a work Wxh.\nBut in inclined plane we move the effort P through a distance l . Work done is Pxl\n But P x l = W x h \nSince l is greater than h, P is less than w.\nTherefore \nAn inclined plane reduces the effort force by increasing the distance through which the force is applied\n\nIn your choise the second and fourth answer is  wrong\nas  no  (ideal) machine can increase or reduce our work. In an inclined plane effort distance (length of the plane) is more than the height. Therefore the first one is correct.
249841
Work equals the vectors of energy.
777663
2t, where t = arctan(Electrostatic force/Weight);\nThe electrostatic force will act horizontal, and the weight vertical, and these two must compose to a force along the suspending thread.\n\nSubstitute the numerical values in and find the angle.
80396
Yeah i know:\nDo this:\n\na.)Find the work done by the applied force.\n\nb.)Find the work done by friction\n\nc.)Find the work done by gravity\n\nd.)Find the work done by the normal force.\n\ne.) IF the object starts at rest, how fast is it going after the 5.0m.\n\nhope this helped\n-----------\n\nNO I WILL NOT HLEP YOU. DO YOUR OWN WORK! your asking someone to do your work for you.