Diagram on right for Problems 12-15: Assume a frictionless pulley.T125aNFN351 kg125 kg351 kgH= 0.200351a45°351(9.8)125(9.8)12. Calculate the force of the friction (FN) on the 351 kg mass.EFy: N -351 (9.8)cos(45°) = 0%3Dcos(45) = sin(45°) = 0.707FN = µN13. Calculate the acceleration a of the 125 kg mass.EFx: 351a = 351(9.8)sin(45) - T -FN125a = -125(9.8)+ T14. Calculate the tension (T) in the cable. (Hint: substitute a one of the above equations andsolve for T) (round answer to three significant digits)a. 171 Nb. 980 Nc. 1255 Nd. 1414 N15. If the 351 kg mass moves for 3 seconds, what distance will it travel? (Use formula sat)Sa. 4.86 mb. 6.82 mc. 9.72 md. 14.1 m

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Diagram on right for Problems 12-15: Assume a frictionless pulley. T N EN 125a 351 kg H= 0.200 351ks 125 kg 351a 450 351(9.8) 125(9.8) 12. Calculate the force of the friction (Fx) on the 351 kg mass. EFy: N-351 (9.8)cos(45°) = 0 FN = UN cos(45°) = sin(45°)= 0.707 %3D 13. Calculate the acceleration a of the 125 kg mass. EFx: 351a = 351(9.8)sin(45º) – T –FN 125a =-125(9.8) +T 14. Calculate the tension (T) in the cable. (Hint: substitute a one of the above equations and solve for T) (round answer to three significant digits) a. 171 N b. 980 N c. 1255 N d. 1414 N

Diagram on right for Problems 12-15: Assume a frictionless pulley. T N EN 125a 351 kg H= 0.200 351ks 125 kg 351a 450 351(9.8) 125(9.8) 12. Calculate the force of the friction (Fx) on the 351 kg mass. EFy: N-351 (9.8)cos(45°) = 0 FN = UN cos(45°) = sin(45°)= 0.707 %3D 13. Calculate the acceleration a of the 125 kg mass. EFx: 351a = 351(9.8)sin(45º) – T –FN 125a =-125(9.8) +T 14. Calculate the tension (T) in the cable. (Hint: substitute a one of the above equations and solve for T) (round answer to three significant digits) a. 171 N b. 980 N c. 1255 N d. 1414 N

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

Chapter6: Energy Of A System

Section: Chapter Questions

Problem 9OQ

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