Solutions 5

# Solutions 5 - 4.12 IDENTIFY Apply SET UP Let x be in the...

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4.12. IDENTIFY: Apply m F =a r r . Then use a constant acceleration equation to relate the kinematic quantities. SET UP: Let x + be in the direction of the force. EXECUTE: (a) 2 / (140 N)/(32.5 kg) 4.31 m/s . xx aFm == = (b) 2 1 00 2 x x x xv t a t −= + . With 2 1 0 2 0, 215 m x vx a t = . (c) 0 x vv a t =+ . With 0 2 / 43.0 m/s x vv a t x t . EVALUATE: The acceleration connects the motion to the forces. 4.24. IDENTIFY: The reaction forces in Newton’s third law are always between a pair of objects. In Newton’s second law all the forces act on a single object. SET UP: Let y + be downward. / mwg = . EXECUTE: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger’s weight is the gravitational force that the passenger exerts on the earth, upward and also of magnitude 650 N. y y F a m = gives 2 2 650 N 620 N 0.452 m/s (650 N)/(9.80 m/s ) y a . The passenger’s acceleration is 2 0.452 m/s , downward.

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Solutions 5 - 4.12 IDENTIFY Apply SET UP Let x be in the...

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