Question 16, Physics - equation sheet attached Physics 114 Equation SheetConstants and ConversionsKinematics Continuedg = 9.80 m/sFree-fall accelerationΔνInstantaneousainst. = limAt-o AtAcceleration1N = 1 kg m/s?NewtonUniform motion(v) = (v); = constantPosition in uniformX = x + (v)AtMathematics, Scaling and Vectorsb = a* + loga (b) = xmotionLogarithmConstant(v); = (v,); + azAtacceleration:1log(ab) = log(a) + log (b)x, = x, + (v,),At +a, (at)?log Ax" = n log x + log A(v,); = (v,)} + 2a,AxVolume of a sphereV =Surface area of aA = 4ar?ForcessphereNewton's second lawFnet = EF = mã%3DVolume of a cylinderV = arlNewton's second lawFnetx = EF = ma,%3DSurface area of aA = 2ar? + 2rlComponent formFnety = ER, = maycylinderMass densityp = m/VNewton's Third LawFA en =-ton AA, = A cos e (rel. to x-axis)Weightw = mgx -component of avector ÅApparent weightWapp = magnitude of supporting forcesy -component of aAy = A sin 8 (rel to x-axis)vector ÅKinetic frictionfk = HanMagnitude of vector ẢStatic frictionA = JA + A,Reynolds numberRe = pvl/nDirection of A relative8 = tan-(Ay/A,)1Drag (high Reynoldsnumber)=CopAv?to x-axisAddition of two vectors If = Å + B, thenC, = A, + B,D = 6nyrvDrag (low Reynoldsnumber)Cy = Ay + ByCircular MotionKinematicsCentripetal accelerationa =DisplacementAx = x - XAverage VelocityAxFrequency1VargT2arAtAxVinst. = limInstantaneous VelocityAt+0 AtAvAverage AccelerationdavgΔε A box of mass 60 kg has two horizontal forces acting on it. The first force pushes from the left andhas magnitude F1 = 130 N. The other force drags the box from the right and has magnitude F2. Theground is rough and horizontal, and the box speeds up with an acceleration of 3.7 m/s? in thedirection of the applied forces.If the magnitude of the force of friction is 50 N, what is the value of F2? (Arrows representing forcesare not drawn to scale).O 170 NO 220 NO 80 NO 140 NO 90 N

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Description: Question 16, Physics - equation sheet attached Physics 114 Equation SheetConstants and ConversionsKinematics Continuedg = 9.80 m/sFree-fall accelerationΔνInstantaneousainst. = limAt-o AtAcceleration1N = 1 kg m/s?NewtonUniform motion(v) = (v); = const

mass = m = 60 kg acceleration = a = 3.7 m/s2 F1=130 N frictional force = f = 50 N F2= ?

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A box of mass 60 kg has two horizontal forces acting on it. The first force pushes from the left and has magnitude F1 = 130 N. The other force drags the box from the right and has magnitude F2. The ground is rough and horizontal, and the box speeds up with an acceleration of 3.7 m/s? in the direction of the applied forces. If the magnitude of the force of friction is 50 N, what is the value of F2? (Arrows representing forces are not drawn to scale).

A box of mass 60 kg has two horizontal forces acting on it. The first force pushes from the left and has magnitude F1 = 130 N. The other force drags the box from the right and has magnitude F2. The ground is rough and horizontal, and the box speeds up with an acceleration of 3.7 m/s? in the direction of the applied forces. If the magnitude of the force of friction is 50 N, what is the value of F2? (Arrows representing forces are not drawn to scale).

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter4: The Laws Of Motion

Section: Chapter Questions

Problem 33P: Two blocks, each of mass m = 3.50 kg, are hung from the ceiling of an elevator as in Figure P4.33....

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