a 3.5 kg block is pushed along a horizontal force...?
964900
If i understand your scenario, the force is pushing downward on the block.  The horizontal component of the force is 15 N times cosine of 40 degrees.  The vertical component of the force is 15 N times sine of 40 degrees.  However, there is also the force of gravity, which is (3.5 kg * 9.8 m/s2)  (units are N, like the other forces).  So if you add the force of gravity to the vertical component, you have the total downward force.  Multiply this by the coeficient of friction, and you have the force of friction.  \n\nThe force of friction is directed against the direction of motion.  Subtract the frictional force from the horizontal component, and you have the total horizontal force.  Divide the horizontal force by the mass and you should get the acceleration.\n\nfriction force = [(15*sine 40)+(3.5*9.8)]*.25\nacceleration = [(15*cosine 40)-friction force]/3.5
833368
find the force of her weight. \nthat should be the answer
657765
1. the object is pushed up by the weight of its equivalent volume of water, 4-3.2=0.8l\nSo the object's density is equal to its mass divided by its volume:\n4kg/0.8l=4kg/0.0008m3=40,000kg/8m3=5000kg/m3\n\n2. the force decreases with the square of distance, so it will be 3^2=9 times smaller, which is 6N.
40451
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)
1093836
Centrifugal force and gravity.
962088
Because the force is pulled at an angle to the bank.  The force has an x and y component.  The x component (parallel to bank) will be 490.3.  Get this by solving the right triangle.  Hyp = 10.  Opp. = 2.\n\nWith that force the initial acceleration is 0.098 m/s^2.
744947
First you should draw a free-body diagram to understand the forces acting on the block.  In this case, you have 4 forces:\n\n1) 40 N force moving in the +x direction, let's denote it F_1\n2) frictional force moving in the -x direction, let's denote it F_2\n3) Weight, moving in the -y direction, let's denote it W\n4) Normal force opposing weight, let's denote it N\n\nNow you can write Newton's 2nd law equations:\n\nx-direction: F_1 - F_2 = m*a\ny-direction N - W = 0 (equal to zero because block does not move in this direction)\n\nUsing these equations, you know that N = W = m*g\n\nUsing F_2 = u*N and a = 6 m/s^2m, we can solve for thd desired variables:\n\na)\nF_2 = F_1 - m*a  ---------> 40 - (5)(6) ---------> F_2 = 10 N\n\nb)\nF_2 = u*N = u*m*g, now solve for u:\n\nu = F_2/(m*g) = (10)/(5*9.8) -----------> u = 0.204\n\n--------------\n\nHope this helps
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.
481598
There are three objects; 5kg, 9kg, and a cable of negligible mass.\n\nWhen the three move, they must have the SAME acceleration.\n\nTherefore, consider the three as a single object of mass 5 + 9 = 14 kg, with acceleration ‘a’.\n\nThe only force that is pulling the three is the force given by the 9kg object. (The gravitational pull is only on this 9kg.)\n\nThis force is 9 x 9.8 N = F.\n\nSince F = M a,   9 x 9.8 = 14 x a.\n\nFrom this a = 6.3 m/s^2.\n\nAll the three objects are moving with an acceleration 6.3 m/s^2.\n\nIf the body 5kg is moving with an acceleration of 6.3 m/s^2, then what is the force acting on it?\n\nWe know that f = ma; hence, f = 5 x 6.3 = 31.5 Newton.\n\nSince the tension of the rope is the pulling force, tension =31.5 Newton.\n\nThe tension can be calculated using the forces acting on the other end of the cable also.\n\nThe downward force on 9kg is 9 x 9.8 N; the tension is the upward force T. \n\nThe net force acting on 9kg is (9 x 9.8 - 9 x 6.3) = 9 x 3.5 = 31.5 N = T.
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.
965191
This isn't very hard - you just need to do each of the little pieces. I'll give some hints.\n1) When the block is at the extreme from the rest position all the system energy is stored in the spring. The stored energy is spring force (k*x) times spring deflection distance (x). So it's k*x^2 where x is 8.5. \n\nThe block has maximum speed when the block is in the center of the oscillation range (the spring is fully relaxed at that point), so all the energy is in kinetic energy. Energy is conserved, so\nMax spring energy = Max KE\nk*x^2 = (1/2)*m*v^2\n\n2) Best hint here: think, no math req'd.\n\n3) Spring force F = k*x. Acceleration = F/m\n\n4) Acceleration is max (see #3) at x = 8.5 and zero at x = 0. In the formula\nx = (1/2)*a*t^2,\nyou can use the average of the accelerations those 2 points for a.
1036294
letter b is the answer.. because the weight of the object is approx. 627.20kg multiplied by 9.8(acceleration due to gravity) so the product is the force..
730789
The force pulling the block down is m*g*sin(ø);  the force of friction is k*m*g*cos(ø),  The net force is the m*g*sin(ø) - k*m*g*cos(ø).  The resulting acceleration is this force divided by m, so\n\na = g*sin(ø) - k*g*cos(ø)\n\na = g[sin(ø) - k*cos(ø)]
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
985550
consider the forces on the block:\n\nThe force of gravity\n\nThe frictional force\n\nWhen the forces are translated into the frame of reference of the surface of the slope,\nthe forces parallel to the slope are:\n\nfriction:\n0.60*cos(60)*m*g\nWeight\nsin(60)*m*g\n\nnote that in order for the block to be sliding down the slope, the parallel component of the weight must be greater than the frictional force \nor sin(60) must be greater than .6*cos(60)\nwhich it is\n\nTo compute the acceleration,\nF=m*a\nwhich is\nm*a=sin(60)*m*g-0.60*cos(60)*m*g\ndivide out the mass\na=g*(sin(60)-0.60*cos(60))\na=5.55 m/s^2\n\nb)  For this part, only the force necessary to overcome kinetic friction plus the parallel weight is necessary to keep it moving up the slope.  Since it was given a shove and we don't know how much force was applied in the shove, we can only compute the minimum force necessary to over come kinetic friction and weight\nF>g*m*(sin(60)+0.60*cos(60))\n\nj
770512
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.