# ch6 - 6.3.IDENTIFY Each force can be used in the relation W...

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6.3.IDENTIFY: Each force can be used in the relation (c o s) WF s F s φ == & for parts (b) through (d). For part (e), apply the net work relation as net worker grav . nf WW W W W =+ + + SET UP: In order to move the crate at constant velocity, the worker must apply a force that equals the force of friction, worker k k . Ffn μ EXECUTE: (a) The magnitude of the force the worker must apply is: () ( ) 2 worker k k k 0.25 30.0 kg 9.80 m/s 74 N Ff n m g μμ == = = = (b) Since the force applied by the worker is horizontal and in the direction of the displacement, 0 = ° and the work is: ( )( ) [ ]( ) worker worker cos 74 N cos0 4.5 m 333 J s = ° + (c) Friction acts in the direction opposite of motion, thus and the work of friction is: 180 = ° ( ) ( ) [ ]( ) k cos 74 N cos180 4.5 m 333 J f Wf s = ° (d) Both gravity and the normal force act perpendicular to the direction of displacement. Thus, neither force does any work on the crate and grav 0.0 J. n (e) Substituting into the net work relation, the net work done on the crate is: net worker grav 333 J 0.0 J 0.0 J 333 J 0.0 J W W W + + = + + + = E VALUATE : The net work done on the crate is zero because the two contributing forces, are equal in magnitude and opposite in direction. worker and , f FF 6.4.IDENTIFY: The forces are constant so Eq.(6.2) can be used to calculate the work. Constant speed implies We must use 0. a = m = F a G G applied to the crate to find the forces acting on it. (a) SET UP: The free-body diagram for the crate is given in Figure 6.4. EXECUTE: y y Fm a = sin30 0 nm gF −° = sin30 = kk k k sin30 fn m g F = ° Figure 6.4 x x a = k cos30 0 °− = cos30 sin30 0 g ° = F 2 k k 0.25(30.0 kg)(9.80 m/s ) 99.2 N cos30 sin30 cos30 (0.25)sin30 mg F ° ° = (b) ( cos ) (99.2 N)(cos30 )(4.5 m) 387 J F s ° = ( cos30 F ° is the horizontal component of ; F G the work done by F G is the displacement times the component of F G in the direction of the displacement.) (c) We have an expression for k f from part (a): 2 ( sin30 ) (0.250)[(30.0 kg)(9.80 m/s ) (99.2 N)(sin30 )] 85.9 N fm g F ° = + ° = 180 since k f is opposite to the displacement. Thus k ( cos ) (85.9 N)(cos180 )(4.5 m) 387 J f s ° =

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(d) The normal force is perpendicular to the displacement so 90 φ = ° and 0. n W = The gravity force (the weight) is perpendicular to the displacement so 90 = ° and 0 w W = (e) tot 387 J ( 387 J) 0 Ffnw WW W W W =+++= + + = E VALUATE : Forces with a component in the direction of the displacement do positive work, forces opposite to the displacement do negative work and forces perpendicular to the displacement do zero work. The total work, obtained as the sum of the work done by each force, equals the work done by the net force. In this problem, net 0 F = since and which agrees with the sum calculated in part (e). 0 a = tot 0, W = 6.17.IDENTIFY and SET UP: Apply Eq.(6.6) to the box. Let point 1 be at the bottom of the incline and let point 2 be at the skier. Work is done by gravity and by friction. Solve for 1 K and from that obtain the required initial speed.
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## This note was uploaded on 04/29/2008 for the course PH 1110 taught by Professor Kiel during the Fall '08 term at WPI.

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ch6 - 6.3.IDENTIFY Each force can be used in the relation W...

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