what is the acceleration of a 40kg block of cement when pulled sideways with a net force of 200N.  if a mass of 1kg is accelerated 1m&#92;s2 by a force of 1N what would be the aceleration of 2kg acted on by a force of 2N.  two boxes are seen to accelerate at the same rate when a force F is applied to the first and 4F is applied to the second, what is the mass ratio of the boxes?
215948
the answer to the first question is 5 m/s.\nF=ma; a=F/m=200N/40kg=5m/s\nthe second answer is 1m/s.\nF=ma; a=F/m=2N/2kg=1m/s\nthe third answer is 1:4.\nF=ma; m=F/a\n(1) M=f/a=F/a=m\n(2) M=f/a=4F/a=4m
75670
look at the question and it provides the answer --- it is acceleration occurring at a uniform rate (ie. it is constant).  for example, from a stationary start, a uniform acceleration of 2m/s^2 will give speeds of 0s=0m/s, 1s=2m/s, 2s=4m/s, 3s=6m/s.
538354
Well it has a mass of 27.4 kg\nYou are pushing with a force of 10.5 N\n\nSince the mass does not change, the acceleration = Force/Mass so,\n\na = 10.5/27.4 m/s^2\n\nthis is a constant value and therefore x = 0.5 * a * t^2\n\nso x = 0.5 * 10.5  * 2.03 * 2.03 / 27.4 meters\n\nyou can multiply it out
527409
This problem relies on Newton's second law: F=ma\n\na. This question is not completely specified.  The mass of the rocket is then (F/a=m or 2.0E7/a).\n\nb. We first find the net force on the rocket and use the mass from the previous question. Now we compute the acceleration at liftoff (F/m=a or (25E6 - 2.0E7)/(2.0E7/a)).\n\nc. We derive a new formula by double integration on acceleration. \nintegrate(a dt) = at, integrate(at dt) = .5at^2. The sequence gives us acceleration -> speed -> displacement at time t. The rocket will be at velocity: +(25E6 - 2.0E7)/(2.0E7/a)(7*60) m/s and at displacement of: +.5((25E6 - 2.0E7)/(2.0E7/a))(7*60)^2 m.
527034
Here you go.\nNewton's second law of motion can be formally stated as follows:\n\nThe acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.\nIn terms of an equation, the net force is equated to the product of the mass times the acceleration.\n\nFnet = m * a\nIn this entire discussion, the emphasis has been on the "net force." The acceleration is directly proportional to the "net force;" the "net force" equals mass times acceleration; the acceleration in the same direction as the "net force;" an acceleration is produced by a "net force." The NET FORCE. It is important to remember this distinction. Do not use the value of merely "any 'ole force" in the above equation; it is the net force which is related to acceleration. As discussed in an earlier lesson, the net force is the vector sum of all the forces. If all the individual forces acting upon an object are known, then the net force can be determined. If necessary, review this principle by returning to the practice questions in Lesson 2.\n\n\nThe above equation also indicates that a unit of force is equal to a unit of mass times a unit of acceleration. By substituting standard metric units for force, mass, and acceleration into the above equation, the following unit equivalency can be written.\n\n\nThe definition of the standard metric unit of force is stated by the above equation. One Newton is defined as the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s.\n\n\nThe Fnet = m a equation can also be used as a "recipe" for algebraic problem-solving. The table below can be filled by substituting into the equation and solving for the unknown quantity. Try it yourself and then use the "pop-up menus" to view the answers.\n\n\n Net Force\n(N) Mass \n(kg) Acceleration\n(m/s/s) \n1. 10 2  Depress mouse.a = 5 m/s/sa = F/ma = 10 N/2 kga = 5 m/s/s \n2. 20 2  Depress mouse.a = 10 m/s/sa = F/ma = 20 N/2 kga = 10 m/s/s \n3. 20 4  Depress mouse.a = 5 m/s/sa = F/ma = 20 N/4 kga = 5 m/s/s \n4.  Depress mouse.F = 10 NF = m aF = 2 kg x 5 m/s/sF = 10 N 2 5 \n5. 10  Depress mouse.m = 1 kgm = F/am = 10 N/10 m/s/sm = 1 kg 10 \n\n\nThe numerical information in the table above demonstrates some important qualitative relationships between force, mass, and acceleration. Comparing the values in rows 1 and 2, it can be seen that a doubling of the net force results in a doubling of the acceleration (if mass is held constant). Similarly, comparing the values in rows 2 and 4 demonstrates that a "halving" of the net force results in a "halving" of the acceleration (if mass is held constant). Acceleration is directly proportional to net force.\n\nFurthermore, the qualitative relationship between mass and acceleration can be seen by a comparison of the numerical values in the above table. Observe from rows 2 and 3 that a doubling of the mass results in a "halving" of the acceleration (if force is held constant). And similarly, rows 4 and 5 show that a "halving" of the mass results in a doubling of the acceleration (if force is held constant). Acceleration is inversely proportional to mass.\n\nThe analysis of the table data illustrates that an equation such as Fnet = m*a can be a guide to thinking about how a variation in one quantity might effect another quantity. Whatever alteration is made of the net force, the same change will occur with the acceleration. Double, triple or quadruple the net force, and the acceleration will do the same. On the other hand, whatever alteration is made of the mass, the opposite or inverse change will occur with the acceleration. Double, triple or quadruple the mass, and the acceleration will be one-half, one-third or one-fourth its original value.\n\n \n\nAs stated above, the dire
459782
Fnet = Net Force, a = acceleration, m = mass\n\nFnet = a * m\n\n1) Force = 80m/s2 * 0.4kg = 32N\n\nm = Fnet/a\n\n2) m = 330N/5500m/s2 = 0.06Kg\n\na = Fnet/m\n\n3) a = 220N/55Kg = 4m/s2
552670
Assume superman's mass = m \n\nFriction force developed by shoes: (g = 9.8 m/s^2)\nFf = m * g * 0.9\n\nAcceleration from force:\nF = m * a\n\nEquate the forces such that F = Ff, solve for acceleration:\nm * a = m * g * 0.9\na = 8.82 (m/s^2)\n\nDetermine the time required to go 50 m:\nx = (1/2) * a * t^2\n50 = 0.5 * 8.82 * t^2\nt = 3.367 seconds\n\nTada!
406941
Acceleration and Force are directly related. So the question is the same as asking for the position that gives 1/2 max force.\n\nSpring force and displacement are directly related. So the question is the same as asking for the displacement that is1/2 max displacement.
183202
force = m*a + I*alpha\n\nthe first part is for the cart and the second is for the pulley\n\nThen relate the angular acceleration to the linear acceleration as follows:\nalpha=angular acceleration= w^2*r   in rad/s\nso a(accleration of cart)= w^2*r^2   m/s\n\nso force=m*a + I*(a/r)\n\nfrom which you can find a (acceleration)
652997
Net force on object = Force applied - Friction force\n\nSince velocity is constant, the acceleration of the object is zero\n\nSo, Net force = mass * acceleration = zero\n\n0 = Force applied - Friction force\nor\nForce applied = friction force \n= coefficient of friction * Normal force\n=.250 * 11.9 kg * 9.8 m/s^2\n=29.2 kg-m/s^2\n\nBox is decelerating at .250*9.8m/s^2\n\nand\n\ndistance traveled = (final velocity^2 - initial velocity^2)/2a\n=(0 - 3.2 m/s)  /  2*(-.250*9.8m/s^2)\n=.65 m
240668
Force = mass x acceleration\n\nor \n\nAcceleration = force / mass\n\nIf mass is tripled and force is halved, the acceleration drops to 1/6 its initial value.
500313
a=F/m\n\nF=30N\nm=60kg\n\na=30/60\na=.5m/s^2
30699
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
667296
Let us write coeff. of friction = k     \n                    acceleration due to gravity = g     m/ sq. s\n                    uniform velocity of motion = u      m/s\n                 \n                    Since k=F/R where R = weight of the box acting downward and F = Frictional force acting against the direction of motion.\n                   Hence, F = kR. Again F= ma, where a is the retardation on the box and R=mg.\n                    Therefore, a=kg. (m cancels out on both sides).\n\n                     By using the formula:\n                            v^2=u^2 - 2aS  ( here a is retardation) we have\n                      S= u^2/(2a) = u^2/(2kg).\n\n            Substituting the values:\n                         u=8 m/s\n                         k= 0.16\n                         g=9.8 m/sq. s\n                        The result is S= (8^2)/(2*0.16*9.8) = 20.41 m (approx.).
237173
It's true - gravity pulls harder (i.e. more force) on a more massive (i.e. heavier) object. BUT that object is harder to accelerate - it has more interia. The two effects cancel each other out.\n\nClearly you don't like equations, but putting it in numbers can help:\n\nIf a 1kg object is pulled towards the earth, what will be the force, and what will be the acceleration?\nForce = mass * earth's gravitational field strength\nso Force = 1kg * 9.81N/kg\nForce = 9.81N (or 10N, roughly)\n\nAcceleration = Force / Mass\nSo Acc'n = 9.81N / 1kg = 9.81 m/s^2\n\nNow consider the same for a 10kg object:\n\nForce is now 98.1N (or 100N, roughly)\nSo Acc'n = 98.1N / 10kg = 9.81 m/s^2\n\nSo the acceleration is the same.
415870
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
484301
Get the unbalanced force from the first 2 #s in your ? :  2.8-2.6 = .2\n\nb)  F = ma---> m = F/a = 0.2/.11 = 1.818 kg