ch4 - Physics 101 Chapter A Homewor where g changes with...

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Unformatted text preview: Physics 101 Chapter A]. Homewor where g changes with location. «0 Wm = mgm = (76 kg)(9-8m/s‘) = (b) Wm = mgMM =(76 kg)(1.7 m/s‘) = (c) Wm = "1ng = (76 kg)(3.7 m/s’) = (d) WSW = mgsw = (76 kg)(0m/s2) = 4. Inallcases, W =mg, Find the average second law. 6. v=0 vo FM The negative sign ' =mam 17. (a) There will be two forces on the upward force of air resistance, FA Newton’s 2'“1 law for the skydivers. ZF=FA —mg=ma —+ 0.25mg—mg=ma —> a = —0.75g = —o.75_(9.8 m/sz) = Due to the sign of the result, therd (b) If they are descending at constant speed, then be zero, and so the force of air resistance must FA = mg = (132 kg)(9,80 = skydivers — combined weight, and the . Choose ufito be the positive direction. Write irection of the acceleration is down. the net force on them must be equal to their weight. W. Abdul-Razzaq k Solution acceleration from Eq. 2-2. The average force on the car is found from NeWton’s 26. 27. We assume that the mower is being pushed to the right. Ffr is the (a) friction force, and F, is the pushing force along the handle. "Write Newton’s 2nd law for the horizontal direction. The forces must sum to 0 since the mower is not accelerating. ' ZFI =15;,cos45.0°—Ffr =0 —-) Ffi = F], cos45.0° = (88.0 N)cos45.0° = Write Newton’s 2ncl law for the vertical direction. The forces must sum to 0 since the mower is not accelerating in the vertical direction. 21.; =1:N —mg—F,sin45.0° =0 —) FN = mg+F,, sin 45° = (14.0 kg)(9.80.m/s2)+(88.0 N)sin45.0° = First use Eq. 2-1 la to find the acceleration. v—vo _1.5m/s—0 t 2.5 s Now use Newton’s 2"“l law for the x direction to find the necessary pushing force. ZFX =F;, cos45.0° —Ff = ma __) 62.2 N +("14.0 kg)(0.60m/s’) = -99.9 N cos 450° (17) (c) (d) = 0.60 m/sz. v—vo=at —-> a: E + ma FF = o _ cos45.0 Choose the y direction to be the “forward” direction for the motion of the snowcats, and the x direction to be to the right on the diagram in the textbook. Since the housing unit moves in the forward direction on a Straight line, there is no acceleration in the x direction, and so the net force in the x direction must be 0. Write Newton’s 2“d law for the x direction. ZE=FM+FBX=0 —> —FAsin50°+FBsin30°=0 -—) ‘ ° 4500 N ' 50° FAS‘“50 ( _ )m = 6.9x10’N sin 30° sm 30° Since the x components add to 0, the magnitude of the vector sum of the two forces will just be the sum of their y components. ZFy = FM + Fay = FA cos 50° + F5 cos 30° = (4500 N) cos 50° + (6900 N)cos30° = 8.9x 10’N ,2» ' g k . x. »- f . 4- V" i I hi“ w an ‘ (AP M . - " VP 7 E / for the crate is shown. The crate does not accelerate -. = mg . The crate does not accelerate horizontally, and F Lg: f” fr 36. A free-body diagram vertically, and so FN so FF = Ffr , Putting this together, we have = ykmg = (o.30)(35 kg)(9.8m/s2) = 103 = FP=Ffi=flkFN 1f the coefficient of kinetic friction is zero, then the horizontal force required is , since there is no friction to counteract. Of course, it would take a force to START the crate moving, but once it was moving, no further horizontal force would be necessary to maintain the motion. ' rt? body diagram for the carton on the surface. There (81) Consider the free- . . . is no motion in the y direction and thus no acceleration in the y direction. Write Newton’s 2nd law for both directions. ' ZFX = FN —mgcos¢9 = 0 —) 1'4"N = mgcosfi 21:; =mgsint9—Ffr =ma ma = mg sin6i—,u,{FN = mg sinQ—pkmg cos6? a = g(sin6—yk cost?) = (9.80 m/sz)(sin 22.o° —o.12coszz.o°) = 2.58 = (1,) Now use Eq, 2-11c, with an initial velocity of O, to find the final velocity. v2 -1}: = 2a(x-—xo) ——) v=,’2a(x—-xo) =1’2(2.58m/s2)(9.30 m) = 6.9m/s \WJQ free—body diagram for the carton on the fiictionless 53. (a) Consider the . . , . . , surface, There is no acceleration in the y direction. Write Newton 5 2"“ law for the x direction. 2F; =mgsin¢9=ma —> a=gsin6 that it slides before stopping. v2 -v: = 2a(x—xo) —) 2’ 2 _ _ 2 V V0 _ 0 ( = (x-xo) =7 " 2(9.8m/s2)sin22.0° The negative Sign means that the block is displaced up the plane, which is the negative direction. (b) The time for a round tI’ip can be found from Eq. 2—11a. The free-body diagram (and thus the acceleration) is the same whether the block is rising or falling. For the entire trip, v0 = —3.0m/s and v = +3.0m/s. z (3.0m/s>—<—3.0m/s> = = t: v v°+at i) a (9.8m/sz)sin22° _.33..— ...
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This note was uploaded on 10/13/2010 for the course PHYSICS 101 taught by Professor Abdulrazzaq during the Spring '10 term at West Virginia State University.

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ch4 - Physics 101 Chapter A Homewor where g changes with...

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