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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).

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Publisher:Raymond A. Serway, Chris Vuille
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Question 16, Physics - equation sheet attached

Physics 114 Equation Sheet Constants and Conversions Kinematics Continued g = 9.80 m/s Free-fall acceleration Δν Instantaneous ainst. = lim At-o At Acceleration 1N = 1 kg m/s? Newton Uniform motion (v) = (v); = constant Position in uniform X = x + (v)At Mathematics, Scaling and Vectors b = a* + loga (b) = x motion Logarithm Constant (v); = (v,); + azAt acceleration: 1 log(ab) = log(a) + log (b) x, = x, + (v,),At +a, (at)? log Ax" = n log x + log A (v,); = (v,)} + 2a,Ax Volume of a sphere V = Surface area of a A = 4ar? Forces sphere Newton's second law Fnet = EF = mã %3D Volume of a cylinder V = arl Newton's second law Fnetx = EF = ma, %3D Surface area of a A = 2ar? + 2rl Component form Fnety = ER, = may cylinder Mass density p = m/V Newton's Third Law FA en =- ton A A, = A cos e (rel. to x-axis) Weight w = mg x -component of a vector Å Apparent weight Wapp = magnitude of supporting forces y -component of a Ay = A sin 8 (rel to x-axis) vector Å Kinetic friction fk = Han Magnitude of vector Ả Static friction A = JA + A, Reynolds number Re = pvl/n Direction of A relative 8 = tan-(Ay/A,) 1 Drag (high Reynolds number) =CopAv? to x-axis Addition of two vectors If = Å + B, then C, = A, + B, D = 6nyrv Drag (low Reynolds number) Cy = Ay + By Circular Motion Kinematics Centripetal acceleration a = Displacement Ax = x - X Average Velocity Ax Frequency 1 Varg T 2ar At Ax Vinst. = lim Instantaneous Velocity At+0 At Av Average Acceleration davg Δε
Transcribed Image Text:Physics 114 Equation Sheet Constants and Conversions Kinematics Continued g = 9.80 m/s Free-fall acceleration Δν Instantaneous ainst. = lim At-o At Acceleration 1N = 1 kg m/s? Newton Uniform motion (v) = (v); = constant Position in uniform X = x + (v)At Mathematics, Scaling and Vectors b = a* + loga (b) = x motion Logarithm Constant (v); = (v,); + azAt acceleration: 1 log(ab) = log(a) + log (b) x, = x, + (v,),At +a, (at)? log Ax" = n log x + log A (v,); = (v,)} + 2a,Ax Volume of a sphere V = Surface area of a A = 4ar? Forces sphere Newton's second law Fnet = EF = mã %3D Volume of a cylinder V = arl Newton's second law Fnetx = EF = ma, %3D Surface area of a A = 2ar? + 2rl Component form Fnety = ER, = may cylinder Mass density p = m/V Newton's Third Law FA en =- ton A A, = A cos e (rel. to x-axis) Weight w = mg x -component of a vector Å Apparent weight Wapp = magnitude of supporting forces y -component of a Ay = A sin 8 (rel to x-axis) vector Å Kinetic friction fk = Han Magnitude of vector Ả Static friction A = JA + A, Reynolds number Re = pvl/n Direction of A relative 8 = tan-(Ay/A,) 1 Drag (high Reynolds number) =CopAv? to x-axis Addition of two vectors If = Å + B, then C, = A, + B, D = 6nyrv Drag (low Reynolds number) Cy = Ay + By Circular Motion Kinematics Centripetal acceleration a = Displacement Ax = x - X Average Velocity Ax Frequency 1 Varg T 2ar At Ax Vinst. = lim Instantaneous Velocity At+0 At Av Average Acceleration davg Δε
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). O 170 N O 220 N O 80 N O 140 N O 90 N
Transcribed Image Text: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). O 170 N O 220 N O 80 N O 140 N O 90 N
Expert Solution
Step 1

mass = m = 60 kg

acceleration = a = 3.7 m/s2

F1=130 N

frictional force = f = 50 N

F2= ?

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