A mover loads a 253 kg crate of hammers onto a truck.  If 3650 of work are required to raise the crate from the pavement to the truck bed how far is the crate lifted?\n\nusing equations f=ma and w=fd, how would i solve that? and what's the answer?
132613
well, using f=ma, and w=fd, we immediately get w=(ma)d\nW=work  (3650 joules)\nM=mass  257kg\na=accelleration =gravity =9.81\nd=distance (required)\n\nrearrange w=(ma)d\n\nd=w/(ma)\nDistance=work/(mass*acceleration)\n\n= 3650/(253*9.81)\n=1.47m
287058
I always find it easier if you draw it first, which this stupid bos wont let me do.\n\nYou know one angle (theta) and one side of the triangle on parts a and b.  Use Tan Theta = x / 30 to find how far the truck has traveled in one second, and convert to MPH.\n\nFor C, you have the distance traveled in MPH and one side (30 ft).  Convert 55 MPH to ft / sec and find the angle Theta.  Tan Theta = 55 MPH / 30 ft\n\nYou are going to have to figure the answers out for yourself.  Good luck
548811
A.momentum is velocity times mass. Here this is 30*30'000=900'000 kg*m/s\n\nB.multiply m/s by 3.6 to get km/h, you have 30*3.6=108km/h\n\nC.distance = velocity * time, here this is 30 * 0.75 = 22.5 meters\n\nD.60 km/h divided by 3.6 gives you 16.67 m/s. Over 0.75s the driver would have travelled just 12.5 meters, i.e. 10 meters less.  So of course it is safer to driver more slowly.  But the catch is, with this reasoning you can show that it is even safer to drive at 40km/h, then 20km/h, and in the end you'll find that remaining at rest is really safe.  But at the end of the day, you still want the truck to get the stuff it transports somewhere, and if possible within a reasonable amount of time... ;-)\n\nhope this helps
15433
well figure out the net force with F=ma and just subtract. like just say 93 x .26 is the net force, the force thats acually being used for motion. so you subtract this from 36 to see what force is opposing this. so the total equation is:\n36 - (93 x .26)
346168
the constant velocity indicates that the 30.0 N force is just enough to overcome the force of friction\n\nSo use 30 N in the friction formula:\n\nmu=Force of Friction / Normal Force , Normal Force = Weight of Crate\n\nmu = 30.0/(9.81 x 12 ) = 0.2548419979612640163098878695209
32900
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
25002
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 ; )
222354
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
173281
First put everything in the same units.\n\n36 km/h * 1000m/1km * 1hr/3600s = 10m/s\n\nOk...the equation for on objects position as a function of time is.\n\nx(t) = xo + v*t + a*t^2\n\nwhere \nxo = initial position (0 for both car and truck)\nv  = velocity (10 m/s for the truck)\na  = acceleration (4 m/s2 for the car)\nt  = time\nx(t) = position at any given time t\n\nIn question 1 you are looking for the time at which the car and the truck are at the same position.  Or the time at which x(t) for the car equals x(t) for the truck.  So set the position equations equal to each other and solve for t.\n\ncar equation:  x(t)car = 0 + 0 * t + 4 * t^2\ntruck equation: x(t)truck = 0 + 10 * t + 0 * t^2\n\ntherefore when they are at the same position\n\n4t^2 = 10t\n\nor\n\nt = 10/4s or 2.5s (note: the answer you have given is incorrect.  At 5 seconds the car would be 100 meters from the light and the truck would be 50 meters from the light)\n\nFor question 2 we need to take the derivative or the position equation for the car and solve for the velocity at time t = 2.5\n\nx(t)car = 4t^2\n\nd(x(t)car)/dt = v(t)car = 8t\n\nv(t)car (at t = 2.5) = 8*2.5 = 20 m/s  your answer for this question proves to be correct
511400
Since the vehicles lock together and travel at 45deg to the original motion of both vehicles, then the momentum of each before the collision must have been equal. We will call the mass of the Truck, Mt and the mass of the Car, Mc. The velocity of the Truck, Vt and the velocity of the car Vc. So:\n\nMtVt = McVc ...........(1)\n\nand \n\nThe total momentum immediately before the collision will be the same as the total momentum immediately after the collision:\n\nMtVt + McVc = (Mt+Mc)Vd ............(2)\n\nwhere Vd is the velocity of the Truck and Car combination immediately after the collision.\n\nAlso, we know that the kinetic energy of the Truck and Car combination immediately after the collision is:\n\n0.5(Mt+Mc)Vd^2................(3)\n\nNow all of this energy is consumed by the work done by the friction between the tires and the road over 18m. So:\n\n0.5(Mt+Mc)Vd^2 = W = Fx18....................(4)\n\nWhere F = (Mt+Mc)xa.....................(5)\n\nand a is the acceleration due to friction = 0.923x9.8 = 9.0454ms^-2\n\nNow, you have values for Mt, Mc and a, so you can calculate Vd from equation (4).\n\nFrom equations (1) and (2), and knowing Vd, you can calculate Vt and Vc.
301552
distance = rate times time or\nd = rt\nLet x = rate of the freight train, then\nx+30 = rate of the express train.\nt = 3 hours for both trains\nFor the freight train, d1 = rt\nd1 = x(3) or d1 = 3x\nFor the express train, d2 = rt \nd2 = (x+30)3 or d2 = 3x + 90\nThe sum of the two distances must equal 390\nd1 + d2 = 390\n3x + 3x + 90 = 390\n6x = 300\nx = 50 km per hour for the freight train\n50+30 = 80 km per hour for the express train
485731
First my other friends...Do not talk with our friend like that....\nSecond you should have mass of crate first.But in parametric answer we have:\nk(1)-k(2)=F*L-mg*Cos(b)*s*L-mg*Sin(b)*L\nF=150 N, L=7.46 m, s=0.27, b=28 Degrees and m=Unknown!!!\nEnd...Thank U...Ramin
206136
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
492498
Let y be required distance from the far end.\n\nThe weight is acting at the C.G i.e., at 2 m from the far end.\n\n2/3 of the weight w is acting upward at y and \n\n1/3 of the weight w is acting upward at 4m from the far end.\n\nTaking moments about the far end, \n\n2/3 W y + 1/3 W 4 = W x 2. \n\n Dividing both sides by 2W and multiplying by 3 on both sides\n\ny + 2 = 3 \n\ny = 1m.
345283
Well the first thing to do in a problem like this is extract information from it.\n1.) Weight of Block = 10kg\n2.) Velocity of block(Vi) = 6m/s\n3.) Final velocity of block(Vf) = 0m/s\n4.) Friction acting on block = 20N The friction is always opposite of the direction of the block.\n\nSecond:\nWhat are we calculating? Distance traveled by block! DISTANCE.  We need a formula that will give us distance as an answer! So...we have final velocity, initial velocity, and distance....lets see...how about this one...\n vf^(2) = vi^(2) + 2*a*d \n\nThis equation has Vf, Vi, and D.  \nVf = 0 m/s\nVi = 6 m/s\nD = What we are trying to find.\nThe only thing missing in this equation that we have to find is (a) or acceleration. Hmm.\n\nThere is actually acceleration on  the block.  Friction is acting on the block in the opposite direction; therefore the block is decelerating.  We can use F=ma to find the force of deceleration.\nFriction Force = 20N\nMass = 10kg\nAcceleration = ?\n\nUse the following equation:\nF=ma\n1.) -20N = 10kg * a\n2.) a = -2 m/s (Yes it's negative b/c we are decelerating!)\n\nNow we go back to the original problem!\n vf^(2) = vi^(2) + 2*a*d \n\nThus\n(0^2) = (6^2) + (2)*(-2)*(d)\n0 = 36 - 4d\n-36 = -4d.\nDistance = 9meters! (Make sure to get the units correct!)\nI think I got this right...if not tell me what I did wrong pls\n\n-----------------------------------------------------------------------------------\nThe guy who did it using Energy equations forgot to add one thing....\nmass x (velocity)^2 = Force x distance\nso 10kg * (6m/s)^2 = 20N * x\nso (10 * 36)/20 = 18 meters\n\nThis is another way to do it.  He is using the energy equations to calculate energy in the block. When the block is in motion there is Kinetic Energy in it.  When the block has no motion, there is zero Kinetic Energy in it.  However, the individual that did this messed up the equation as follows:\nKe = (1/2) * Mass * Velocity^2\n\nThus the answer is not 18 but 9 meters!!!!!\nHere is the correction:\n1.) .5*mass x (velocity)^2 = Force x distance\n2.) .5*10kg*(6m/s)^2 = 20N * x distance\n3.) (5 * 36)/20 = distance&#92;\n4.) Distance =9 meters!
446702
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
366261
SRY!! Matt is going to a basketball game out of town. If he could travel 75mph, He would arrive one hour early. If he could travel 45mph he arrives one hour too Late. How far is Matt's trip\n\ndis=75*(time-1)\ndis=45*(time+1)\n\nthen\ndis= 225mile,    time=4hour