Chapter mv2 = J. y (3,4) F d. 100 x. d = 3 2cos sin100 = 6.84J = 4.52J

Transcription

1 Chapter If a Saturn V rocket with an Apollo pacecraft attached ha a combined ma of m =.9x10 5 kg and i to reach a peed of v = 11. km/ = 11. x10 3 m/, how much Kinetic Energy will it have K = 1 mv = J A coin lide over a frictionle plane and acro an xy coordinate ytem from the origin to a point with xy coordinate (3.0m, 4.0m) while a contant force act on it. The force ha magnitude.0n and i directed at a counterclockwie angle of 100 degree form the poitive direction of the x axi How much work i done by the force on the coin during the diplacement. The eaiet way to do thi problem i to write out both the diplacement vector and the force vector in component form and then ue the definition of work. We begin with a picture of what i happening. y (3,4) F d 100 x d = 3ˆ i + 4 ˆ j F = co100 ˆ i + in100 ˆ j W = F d = 3 co in100 = 6.84J = 4.5J Figure 7-9 how three force applied to a greaed trunk that move leftward by 3m over a a frictionle floor. The force magnitude are F1=5.00N, F=9.00N, and F3=3.00N. During the diplacement, (a) what i the net work done on the trunk by the three force and (b) doe the kinetic energy of the trunk increae or decreae?

2 Since there i no motion in the vertical direction, there i no work done by force and component of force in that direction. We need only concern ourelve with the horizontal motion F net x = F co60 F 1 = 9.00N co N = 0.5N The net force i to the left. Thi i in the ame direction a the motion. The work done i therefore poitive. W = F net x d = 0.5N 3m = 1.5J The work done on the trunk i poitive, o the kinetic energy increae In the figure, a horizontal force F a of magnitude 0.0 N i applied to a 3.00 kg pychology book a the book lide a ditance d = 0.500m up a frictionle ramp at angle θ = 30. (a) During the diplacement, what i the net work done on the book by F a, the gravitation force on the book and the normal force on the book? (b) If the book ha zero kinetic energy at the tart of the diplacement, what i it peed at the end of the diplacement. We begin with a drawing N d θ Fa mg We can calculate the work done by each force

3 W N = N d = Nd co90 = 0 W a = F a d = F a d coθ = 0N 0.5m co 30 = 8.66J = mg d = mgd co(90 + θ) = 3.0kg 9.8m / 0.5m co( ) = 7.35J To find the final peed, we can ue the work to find the change in kinetic energy = W N + W a + = 1.31J 1 mv f 1 mv i = 1 mv f 0 v f = m = m / 7.3 In Fig. 7-33, a block of ice lide down a frictionle ram at angle θ = 50 while an ice worker pull on the block (via a rope with a force F r that ha a magnitude of 50 N and i directed up the ramp. A the block lide through the ditance d = 0.50m along the ramp, it kinetic energy increae by 80J. How much greater would it kinetic energy have been if the rope had not been We begin by computing the work done by the worker. The block lide down the incline, while the force exerted by the worker i up the incline. We begin by computing the work done by the force exerted by the worker W w = F d co180 = 50N 0.5m ( 1) = 5J The total work done on the ice i compoed of a piece due to gravity and a the work that we have computed due to the worker force. We alo know that the total work done i the change in the Kinetic energy. W total = W gravity +W w 80J =W gravity + ( 5J) W gravity =105J If there had been no rope force, the work done would be jut the work of gravity. In that cae, the work done would have been 105J, with a final K of 105 J. Thi i 5J greater than with the rope. 7.4 A cave recue team lift an injured pelunker directly upward and out of a inkhole by mean of a motor driven cable. The lift i performed in three tage, each requiring a vertical ditance of 10.0 m: (a) initially tationary pelunker i accelerated to a peed of 5.00 m/; (b) he i then lifted at the contant peed of 5.00 m/; (c) finally he i decelerated to zero peed. How much work i done on the 80 kg recuee by the force lifting him during each tage.

4 Two force act on the recuee: Tenion in the cable and weight. T mg We are intereted in finding the work done by the tenion. We will do thi by calculating the net work done firt. Stage I. (final velocity = 5m/, initial velocity = 0m/) 1 mv f 0 =1000J = = 1000J ( 7840J ) = 8840J Stage II (final velocity = 5m/, initial velocity = 5 m/) 1 mv f 1 mv i = 0J = = 0J ( 7840J ) = 7840J Stage II (final velocity = 5m/, initial velocity = 5 m/)

5 1 mv f 1 mv i = 0J = = 0J ( 7840J ) = 7840J Stage III (final velocity = 0m/, initial velocity = 5 m/) 1 mv f 1 mv i = 0J 1000J = = 1000J ( 7840J ) = 6840J 7.6 During pring emeter at MIT, reident of the parallel building of the Eat Campu dorm battle one another with large catapult that are made with urgical hoe mounted on a window frame. A balloon filled with dyed water i placed in a pouch attached to the hoe, which i tretched through the width of the room. Aume that the tretching of the hoe obey Hooke law with a pring contant of 100 N/m. If the hoe i tretched by 5.00 m and then releaed, how much work doe the force from the hoe do on the balloon in the pouched by the time the hoe reache it relaxed length. The magnitude of the work i W = 1 k x = 1 100N / m (5.00m) = 150J The work done i poitive becaue the force act in the direction of the motion A 5.0 kg block move in a traight line on a horizontal frictionle urface under the influence of a force that varie with poition a hown in Fig How much work i doen by the force a the block move from the origin to x=8m The work done i the area under the curve W =10N m + 1 m 10N + 1 m ( 5N ) = 5J 7.54 A 50 g block i dropped onto a vertical pring with pring contant k=.5n/cm. the block become attached to the pring and the pring compree 1 cm before momentarily topping.

6 While the pring i being compreed, what work i done on the block (a) by it weight, and (b) by the pring force? (c) What i the peed of the block jut before it hit the pring? (d) If the peed i doubled,what i the maximum compreion of the pring. a. The work done by weight i poitive. The force i downward and o i the motion. W w = mgy = 0.5kg 9.8m / 0.1m = 0.94J b) The work done by the pring i negative. The force i upward but the motion i downward. k =.5N cm 100cm 1m = 50N / m W = 1 k x = 1 50N / m (0.1m) = 1.8J c) The net work done equal the change in Kinetic energy. + W = 0.94J 1.8J = 1.506J K f K i = 0 1 mv i = v i = m = ( 1.506J) 0.5 = 3.47m / d) Ue work to determine the final compreion. The velocity i twice the value found in part c. We chooe the negative olution, ince we know that the ma drop below the zero tarting value...

7 v i = 3.47m / = 6.94m / = ΔK = 0 1 mv i = mgy + 1 ky mgy 1 ky = 1 mv i 1 ky mgy 1 mv i = 0 y = +mg ± (mg) 4 1 k ( 1 mv i ) ( 1 k) = +mg ± (mg) + kmv i ) k = +0.5kg 9.8m / ± (0.5kg 9.8m / ) + 50N / m 0.5kg (6.94m / ) 50N / m = 0.099m