a box is dragged up and down a slope of 20degrees to the horizontal.up force equals 4times down force.?
456850
Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
358449
take the rise over the run.\n\nstart at (0,2)\nto get to (4,3) you move 1 up and 4 to the right.\nSo rise=1 run=4\n\nslope is 1/4
947129
a) Force on the box due to max acceleration witout moving the box = the maximum frictional force = 0.24* Normal force = 0.24*Weightof box = 0.24*mg\n\nSo max acceleration = 0.24*g = 2.352m/s^2\ns = .5at^2 = 10.584\n\nb)\nthis is a vector problem\nthe boat needs to be driven towards west inorder to cross exactly to a point at north.\nresultant speed of boat = 140/20 = 7m/s\nso using pythagorian theorem, U^2 = 7^2 +4^2, or U=sqrt(65) = 8.0623\n\nc)displacement = (11i - 5j) - (7i - 8j)  = (4i + 3j) \nforce = (3i + 4j)\nwork done = Fs = (3i + 4j)(4i + 3j)  = 12 + 12=24\n\nSo change in Kinectic energy  = work done  = 24J
912892
20 legs. 4 legs on each.
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
69843
work = force times distance\nW=Fs
878371
The incline is a simple machine that permits less force to be applied to move an object. The work associated with moving an object on an inclined surface, in the absence of friction, is the same as raising the object of weight (m*g) through a distance H.\n\nFor the inclined plane with an inclination angle of 30 degrees, moving the mass (m) over a distance L would raise the object a distance H given by\n\nH = L*sin(30) = 0.5*L\n\nThe work then is\n\nW = m*g*H = m*g*L/2\n\nFor the values given\n\nW = (1.0 kg)*(10 m/s^2)*(0.5 m)/2\n\nW = 10/4 N-m = 2.5 N-m\n\nThe same answer can be found by looking at the net forces on the object and multiplying by the distance traveled.\n\nFu = 10 N (the upslope force)\nFd = m*g*sin(30) = (1 kg)*(10 m/s^2)/2 = 5 N (downslope force)\n\nFn = Fu - Fd = 5 N (Net force)\n\nW = Fn*L = (5 N)*(0.5 m) = 2.5 N-m
397135
The weight of the object is irrelevant unless one is taking into account air resistance. This is because the net force on the object is the force of gravity, so we have ma = -mg, or a = -g, which is all we need to know for the vertical kinematics equations. The only thing we need to know for the horizontal equation is the initial horizontal speed of the package.\nIf, on the other hand, one was concerned with air resistance, one would also need to know approximate air density and cross-sectional area of the object, as well as be able to numerically solve a nonlinear ordinary differential equation (the frictional force is proportional to some function of velocity) so I doubt this is what you wanted.
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
731589
Well To solve this problem, you need to break up forces in x and y components. Once you find both x and y components, you can find magnitude of force and direction ken must pull. \n\nTo keep the bear from moving, some of the forces (in both x and y direction) equal to zero. \n\nSo, sum of the forces in X-direction = 350sin(60) - 400sin(45) - F(in x direction) = 0\nSum of the forces in y-direction = 400cos(45) + 350cos(60) - F(in y direction) = 0\n\nSolve for F(in x direction) and F(in y direction), you will get these: F(in x direction) = 20.266 lb east and F(in y-direction) = 457.84 lb south\nTo find magnitude = ||F||= sqrt((F(x))^2+(F(y))^2) = 458.29 lb\n\nTo find angle= tan(theta) =(F(y))/(F(x)) where theta is the angle\ntheta = arctan (F(y)/F(x))= 87.5 degree east south\n\nSo Ken must pull the rop with force 458.29 lb in the direction East 87.5 degree South to keep the bear from moving. \nI think this is right. Good Luck.!!!
257076
dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.
929181
A simple incline would do.  Imagine placing a 2x6 plank 8 feet long, one end on the ground, the other end on the platform, and you push the 25 pound box up this plank until it reaches the platform. Of course, one has to wonder why the person couldn't just pick up the box and put it down on the platform.
407712
slope = -0.25\nintercept = 1
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
103276
The work done to lift the box to that height is stored in the body as the potential energy. Hence the work done is 400 J.\n\nAs for the height, potential energy is given by m*g*h, and the weight is given by m*g (=200N here).\nTherefore, 200h = 400 => h = 2m.
799229
a. Calculate the horizontal component of the 485 N.\nFh = 485 N*cos35\nCalculate the force of friction.\nFf = 0.57*319 N\nCalculate the net force.\nFn = Fh-Ff\nCalculate the mass of the box.\nW = m*g\nCalculate the acceleration.\nFn = m*a\nCalculate time to move s = 4.00 meters\ns = (1/2)*a*t^2\n\nb. Repeat with uk = 0.75