Physx141MPch5sol

# Physx141MPch5sol - 5.10 Model We assume that the box is a...

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5.10. Model: We assume that the box is a point particle that is acted on only by the tension in the rope and the pull of gravity. Both the forces act along the same vertical line. Visualize: Solve: (a) Since the box is at rest, a y = 0 m/s 2 and the net force on it must be zero: ( )( ) 2 net 0 N 50 kg 9.8 m/s 490 N FTw Twm g =−= ⇒== = = (b) Since the box is rising at a constant speed, again a y = 0 m/s 2 , = 0 N, and net F 490 N Tw = = . (c) The velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since a y = 5.0 , 2 m/s ( )( ) 2 net 50 kg 5.0 m/s 250 N 250 N 250 N 490 N 740 N y FTwm a = = ⇒= += + = (d) The situation is the same as in part (c), except that the rising box is slowing down. Thus a y = –5.0 and we have instead 2 m/s ( )( ) 2 net 50 kg 5.0 m/s 250 N 250 N 250 N 490 N 240 N y a = = + = Assess: For parts (a) and (b) the zero accelerations immediately imply that the box’s weight must be exactly balanced by the upward tension in the rope. For part (c) the tension not only has to support the box’s weight but must also accelerate it upward, hence, T must be greater than w . When the box accelerates downward, the rope need not support the entire weight, hence, T is less than w . 5.14. Model: We assume that the passenger is a particle acted on by only two vertical forces: the downward pull of gravity and the upward force of the elevator floor. Visualize: Please refer to Figure Ex5.14. The graph has three segments corresponding to different conditions: (1) increasing velocity, meaning an upward acceleration, (2) a period of constant upward velocity, and (3) decreasing velocity, indicating a period of deceleration (negative acceleration). Solve: Given the assumptions of our model, we can calculate the acceleration for each segment of the graph and then apply Equation 5.10. The acceleration for the first segment is 2 10 8 m/s 0 m/s 4 m/s 2 s 0 s y vv a tt == = −− () 2 2 app 2 4 m/s 4 1 1 75 kg 9.8 m/s 1 1035 N 9.8 m/s 9.8 y a ww m g g    ⇒=+= + = + =     For the second segment, a y = 0 m/s 2 and the apparent weight is 2 2 app 1 75 kg 9.8 m/s 740 N m g g =+ = For the third segment,

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Dynamics I: Motion Along a Line 5-2 () 2 32 2 2 app 2 0 m/s 8 m/s 2 m/s 10 s 6 s 2 m/s 1 75 kg 9.8 m/s 1 0.2 590 N 9.8 m/s y vv a tt ww == = −−  ⇒=+ = −=   Assess: As expected, the apparent weight is greater than normal when the elevator is accelerating upward and lower than normal when the acceleration is downward. When there is no acceleration the weight is normal. In all three cases the magnitudes are reasonable, given the mass of the passenger and the accelerations of the elevator. 5.15. Model: We assume that the safe is a particle moving only in the x -direction. Since it is sliding during the entire problem, we can use the model of kinetic friction. Visualize: Solve: The safe is in equilibrium, since it’s not accelerating. Thus we can apply Newton’s first law in the vertical and horizontal directions: net B C k k B C 0 N 350 N 385 N 735 N x x F FFF f fFF = + = = + = + = ( )( ) 2 net 0 N 300 kg 9.8 m/s 2940 N y y FF n w n w m g = = = = = = Then, for kinetic friction: k kk k 735 N 0.25 2940 N f
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## This note was uploaded on 05/05/2008 for the course PHYS 141 taught by Professor Staff during the Spring '06 term at Cal Poly.

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Physx141MPch5sol - 5.10 Model We assume that the box is a...

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