a boy pulls a 30kg on the ground at constant speed.if frictional force is20N,how much work is done thru 5metre
354450
WORK IS EQUAL TO ZERO (READ ALL FOR EXPLAINATION)\nmass of body=30kg\nits weight=30kg X 9.8m/s^2\nweight= 294N\n\nBUT HERE WE DON'T HAVE TO DEAL WITH ITS WEIGHT.\n\nAs the body is moving at constant speed, it means that it's not accelerating and thus no force is being applied to it.\n\nOr in other way\n\nthe frictional force which is 20N is canceling the force with which the boy is moving the body. Thus the boy is also exerting a 20N force (20N - x = 0, of course x=20N)\n\nTHE WORK IS ZERO BECAUSE THE FORCE IS ZERO TOO.\n\nsince the frictional force is cancelling the force of the boy.\n\n20N(frictional force)-20N(force applied)=0\nResultant force = 0\n\nFIND WORK NOW\n\nwork =  net force X distance\nwork =  0  X  5m\nwork =  0 joule\nNo work is done\n\nLISTEN\nThe person below me by the name of sblogs is incorrect as I think. Let me prove this by giving some EXAMPLES\nPUSH THE WALL BUT THE WALL WILL NOT MOVE FROM ITS PLACE THOUGH YOU ARE WASTING ENERGY BUT THE WORK DONE IS ZERO.\nBECAUSE THE WALL DOESN'T MOVE THROUGH SOME DISTANCE.\nWORK= FORCE X DISTANCE\nWORK= (FORCE APPLIED ON THE WALL) X (ZERO METERS)\nWORK= F X 0\nWORK= 0 JOULE
746965
You don't need to find the mass if use conservation of energy to solve the problem.\n\nThe kinetic energy of the can initially is:\n1/2*m*v^2\n=.5*m*2^2\n=2*m\nand when it comes to rest it is zero\nThe energy was lost to the work done by friction.\n\nthe force of friction times displacement is the work:\n\nu*g*m*d\n=.2*9.8*m*d\n=19.6*m*d\nnow set the energy of the system equal to zero:\n2*m-19.6*m*d=0\ndivide out the mass and solve for d:\nd=19.6/2\n=9.8m\n\nj
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Use the equation U = 0.5*k*x^2\n\nU = potential energy (joules)\nx = compression distance (meters)\nk = force constant (N/m)\n\nThis gives us the potential energy of the toy.\n\nU = 0.5 * 296 * 0.0122^2 = 0.0220 J\n\nThen calculate the Kinetic Energy:\n\nKE = 0.5 * m * V^2\n\nKE = Kinetic Energy (joules)\nm = Mass (kg)\nV = Velocity (m/s)\n\nKE = 0.5 * 0.153 * 0.299 = 0.0229 J\n\nTotal Energy = 0.0229 + 0.0220 = 0.0449 J\n\nFor part 2, assume total energy has been converted to potential energy and solve for compression distance:\nx = sqrt (2 * U / k) = sqrt (2 * 0.0449 / 296) = 0.0174 m\n\nAmplitude is twice the compression distance (for positive and negative)\nAmplitude = 2 * 0.0174 = 0.0348 meters\n\nFor part 3, assume total energy has been converted to Kinetic energy and solve for speed:\nV = sqrt (2 * KE / m) = sqrt (2 * 0.0449 / 0.153) = 0.766 m/s
1140843
Force with bike=100*sin30*10=500N \nForce without bike= 0.9*500=450N\nThus, walking will save him 50*300=15000 joules.\nBut, don't forget that the bike has a bigger mechanical advantage or force multiplier that reduces the work. Suppose the MA is 4,then the force with bike will be 500/4=125N and hence work will be 15000/4=3750 J only. Therefore, riding is preferable.
882649
First solve for the force of friction between the block and the table using the formula,\n\nf=uR, where f is the force of friction, u the coefficient of static friction, and R is the force normal to the table. R therefore is the weight of the block equal to 31g.\n\nf=0.425*31*9.8\n =129.1N\n\nThe weight of the sand added to the bucket plus the weight of the bucket itself should be just slightly equal to 129.1N so that the system will begin to move. At that point, however, the block will begin to accelerate because the force of friction acting on it will no longer be static friction but kinetic friction, which is lower in magnitude. Thus there will be a net force equal to 129.1 minus the force of kinetic friction on the block. Thus,\n\nNet force, F=129.1-0.320*31*9.8\n                 =129.1-97.2\n                 =31.9N\n\nF=ma, where F is our net force of 31.9N, m the total mass of block plus mass of sand and bucket, and a is the acceleration of the system.\n\nMass of sand and bucket=129.1/9.8\n                                    =13.2kg\nSubstitute known values:\n\n31.9=(31+13.2)a\na=31.9/44.2\n  =0.72m/s^2
69843
work = force times distance\nW=Fs
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E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
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I'm not going to give you the complete answer because you'll learn more by working through it yourself, but I'll give a few pointers.\n\nAny object travelling in a circle at a constant speed has a centripital acceleration that can be calculated from the velocity and the radius of the circle.\n\nNewton tells us that F[net] = ma, so the object must be experiencing a net centripetal force equal to its mass multiplied by its centripetal acceleration.\n\nWhere does this force come from?  In your problem, it sounds as though the only force that could be pulling the box toward the center of the circle (the centripetal force) is the friction between the box and the bed of the truck.  So, when you add up all the (horizontal) forces to get the net force, the olny one is going to be friction.  F[net] = F[friction] \n\nNow all you need to do set F[friction]  equal to the F[net] that you calculated above (remember that F[net] was equal to your centripetal force), and solve for the coefficient of friction (your textbook shold have the formula for static friction, which will include the coefficient of static friction).\n\nGood luck!
35148
Frictional force, F = u*m*g\nHere, F = 300 N, u = ?, m = 80 kg, g = 10 m/s/s (assume)\nTherefore, 300 = u*80*10\n=> u = 3/8
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Fnet=M*A \n\nYou know that the only forces are friction and the applied force of the orderly because friction and the normel force cancel out.\n\nso:\nFo-Ff=M*A\n70N - xN=80Kg * .4m/s^2\n70N - xN=32Kg*M/s^2 =32N\n-xN=32N-70N\nx=38N\n\nThe force of friction is 38.0N in the direction opposite of motion.
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50 N - friction = m*a= 90*0.35 N\nfriction = 50N - 31.5N = 18.5N
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Getting started, have to overcome static friction.  Therefore, force applied = static friction force just before movement\n\nF = umg\nF = 75N\nm = 20kg\ng = gravity\nfind u, plug and play\n\nsame problem for sliding friction
920572
coefficient of kinetic friction = Friction Force / Weight\n\nFriction Force = (coefficient of kinetic friction)x(55x9.81)\n                       = 188.84 N\n\nSo Net Force = 220 -188.84 = 31.16 N\n\nF=mxa,  a=F/m=31.16/55=0.5665 m/s2
984947
W=F*L – work done by F=145N on distance L=7.2m,\nU=k*m*g*cos(a)*L, work wasted on friction, where k*m*g*cos(a) is force of friction, while\nk=0.31, a=11.7deg, g=9.8m/s^2, m=10.3kg.\nY=m*g*L*sin(a) work done to hoist m to height of L*sin(a) meters.\nE0=m*v0^2/2 initial kinetic energy of m, v0=1.3m/s \nNow energy balance: E0+W=U+Y+E1, E1 is final kinetic energy of m E1=m*V^2/2, V is final speed of m. Thus kinetic energy change dE=E1-E0=W-U-Y.\nAs E1=E0+W-U-Y=m*V^2/2, then V=sqrt[2*(E0+W-U-Y)/m], \nI hope you cope with numeric calculations, all are SI units
791910
The work done on an object by a force is the product of the component of the force in the direction of motion and the distance the object moves. The component of the force in the x direction is simply the magnitude of the force times the cosine of the angle. Thus:\n\nW = 100 N * cos 40° * 20.0 m ≈ 1530 J (to 3 significant digits)\n\nEdit: I don't know whether your teacher is complaining about the units or the numbers. A joule IS a newton·meter, so both units are correct. As for the number, I'd say someone made a typo when compiling the answer key. It happens a lot. Ask her to recompute it herself.
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The work done on the car by the net force equals the change on the car's kinetic energy.\n\nWork = Force * distance\nKE = 1/2 mv^2\n\nWe are given a mass for the car and the final velocity, so from this we can calculate the car's final KE,\nKE = 1/2 (2000 kg) * (2 m/s)^2\nKE = 4000 Joules\n\nSince in this case, KE = W\nW = 4000 Joules = F * d\n\nWe know the net force to be the sum of all the forces acting on th car.  Since there is a "forward" force of 1140 Newtons and a "backward" (what we will call 'negative') force acting on the car of 950 N, the net force is 1140 - 950 = 190 Newtons\n\nW = 4000 J = (190 N) * d\nd = 4000 J / 190 N\nd = 21.05 meters
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I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s