A hand H pulls on a taut horizontal rope R (of mass m = 0.002 kg) that is attach

Question

A hand H pulls on a taut horizontal rope R (of mass m = 0.002 kg) that is attached to a block B (of mass M = 5.00 kg). The resulting acceleration a of the rope and block across the frictionless surface has constant magnitude 0.300 m/s2 and is directed in the positive x direction, (a) Identify all the third-law force pairs for the horizontal forces and show how the vectors in each pair are related, (b) What is the magnitude of the force exerted by the rope on the block? (c) What is the magnitude of the force exerted by the block on the rope? (d) What is the magnitude of the force exerted by the hand on the rope?

Explanation / Answer

mass of the rope m = 0.002 kg mass of the block M = 5 kg acceleration a = 0.3 m/s2 a) let force exerted by hand on the rope is FHR let force exerted by rope on the hand is FRH let force exerted by rope on the block is FRB let force exerted by block on the rope is FBR let force exerted by rope on the block is FRB let force exerted by block on the rope is FBR let force exerted by rope on the block is FRB let force exerted by block on the rope is FBR according to Newton's third law of motion, for every force (action) there is a reaction force, that is equal in magnitude, but opposite in direction(F12 = -F21). from this, the pairs of the horizontal forces,      for hand-rope system,   FHR = -FRH and,      for rope-block boundary, FRB = -FBR and,      for rope-block boundary, FRB = -FBR b) from Newton's second law of motion, the magnitude of the force exerted by rope on block is        FRB = Ma                = (5 kg)(0.3 m/s2)                = 1.5 N c) from Newton's third law of motion, the force exerted by block on rope is        FBR = -FRB                = -1.5 N magnitude: FBR = 1.5 N d) from Newton's second law of motion, the magnitude of the force exerted by hand on rope is        FHR = (M+m)a                = (5 kg+0.002 kg)(0.3 m/s2)                = 1.56 N from Newton's third law of motion, the force exerted by block on rope is        FBR = -FRB                = -1.5 N magnitude: FBR = 1.5 N d) from Newton's second law of motion, the magnitude of the force exerted by hand on rope is        FHR = (M+m)a                = (5 kg+0.002 kg)(0.3 m/s2)                = 1.56 N from Newton's second law of motion, the magnitude of the force exerted by hand on rope is        FHR = (M+m)a                = (5 kg+0.002 kg)(0.3 m/s2)                = 1.56 N

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