= v 0 x. / t = 1.75m / s 2.25s = 0.778m / s 2 nd law taking left as positive. net. F x ! F
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1 Multiple choice Problem 1 A 5.-N bos sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the bos sliding to the right at 1.75 m/s and that it stops in 2.25 s with uniform acceleration. The force that friction exerts on this bos closest to: a) 3.97 N b) 49 N c) 5. N d) 8.93 N e) 38.9N Mass of bos m = 5N 9.8m / = 5.1kg f k m v x = 1.75 m/s a x answer a) Problem 2 A professor holds an eraser against a vertical chalkboard by pushing horizontally on it. She pushes with a force that is much greater than is required to hold the eraser. The force of friction exerted by the board on the eraser increases if she: WARNING: The correct answer may surprise you. Think about the amount of static friction in each case. A) pushes with slightly greater force B) pushes with slightly less force C) pushes so her force is slightly downward but has the same magnitude D) stops pushing E) pushes so her force is slightly downward but has the same magnitude The three diagrams below shows the possible scenario. In all cases the normal force is N = cos ( = for A and B). The maximum static friction is f max s = N µ s. We will assume that in all cases, the upward (+ y, as indicated below) vertical force cancels the downward forces so that the eraser do not fall. A and B C E f s +y f s f s Static riction +x cosθ θ Applied sinθ sinθ θ orce cosθ If the eraser is to remain on the wall (not fall) the y-component of the net force must be zero: y net =. The static friction f s < f s mas calculated for all scenarios below. A and B y-component y net = f s = f s = ANSWER: Clearly E C, y-component +y (shown above) y net = f s + sin = f s = sin inal velocity (x) v x = v x = v x a x t,v x = a x = v x / t = 1.75m / s 2.25s =.778m / s 2 nd law taking left as positive net x = f k = ma x = 5.1kg = 3.97N.778m / E, y-component +y (shown above) y net = f s sin = f s = + sin
2 Problem 3 A series of weights connected by massless cords are given an upward acceleration of 4 m/ by a pull P as shown below. A, B and C are the tensions in the connecting cords. The smallest of the three tensions A,B, and C is closest to: a. 483 N pull P b. 621 N c. 196 N d. 276 N 5. kg e. 8. N C B 1. kg 15. kg 2.kg It is obvious that Tension A, T A, between 15.kg and 2. kg mass, is the smallest tension. To solve see free body diagram on 2. kg (m = 2.kg) below: A ANSWER: D T A a = 4. m/ mg Use 2 nd law y net = T A mg = ma T A = m(g + a) = 2kg 13.8m / T A = 276N Kinematics with Calculus Problem 4 Kinematics with Calculus The position of a particle moving in an xy plane is given by r = ( 2t 4 3)î + ( t5 2t), with r in meters and t in seconds. A) find the average velocity and acceleration for the time interval between t = 1s and t = 3s. B) find the velocity and acceleration at t = 1s in unit-vector notation. C) What is the angle between the positive direction of the +x axis and a line tangent (i.e. v ) to the particles path at t = 1 s? Give your answer in the range of (-18 o ; 18 o ). A) r ( t) = xî + y = ( 2t 4 3)î + ( t5 2t) at t = 1s, r ( 1s) = ( 2( 1s) 4 3)î + ( ( 1s)5 2( 1s) ) = î at t = 3s, r ( 3s) = ( 2( 3s) 4 3)î + ( ( 3s)5 2( 3s) ) = 159mî + 237m v avg = x î + y = x( 3s) x( 1s) î + y( 3s) y( 1s) = 8 m î m t t 3s 1s 3s 1s s s
3 v t = v x î + v y = dx at t = 1s, v 1s = 8 1s î + dy = d 2t 4 3 ( 3 )î + 5( 1s)4 2 = 8 m s î + d ( t 5 2t) = ( 8t 3 )î + ( 5t 4 2) at t = 3s, v 1s î + 5( 3s)4 2 = 8( 3s) 3 a avg = v x î + v y = v x 3s t t 3s 1s b) v ( 1s) = ( 8( 1s) 3 )î + 5( 1s)4 2 a( t) = a x î + a y = dv x î + dv y = 216 m s v x ( 1s) = 8 m s = d 8t 3 î + v y ( 3s) v y 1s 3s 1s î + d 5t 4 2 î + 43 m s at t = 1s, a ( 3s) = 24( 1s) 2 î + 2( 1s) 3 = 24 m s î + 2 m 2 s 2 = 14 m î + 2 m = ( 24t 2 )î + ( 2t 3 ) C) v 1s î + 5( 1s)4 2 = 8( 1s) 3 = 8 m s v x > and v y >,1 st 3m quadrant ( 9 < < ). = tan 1 8m = 2.5, = 2.5 Work by Graphical Integration Problem 5 The figure gives the acceleration of a 3. kg particle as an applied force moves it from rest along an x axis from x = to x = 9. m. The scale of the figures vertical axis is set by a s = 5. m. HINT: or part a) Use = ma, and equation on work in the back. or part b) use the equation on power in the back. A) Calculate the work done after the particle move from = m to = 1.m. Repeat for =.5 m to = 1.m. Repeat for the interval = to = 5.m. B) If at x =, the particle is at rest (at v = ), calculate the speed of the particle when it is at = 5.m. W = dx = m adx = m area under the graph or = to = 1.m: [ ] adx = = 1 2 base ( height) = 1 ( 2 ) ( a s ) = 1 ( 2 ) 5 m # = 2.5
4 W = m a dx = 3kg 2.5 = 7.5J or = to = 1.m: Here I will enlarge the figure for clarity adx = = minus 1.m adx = 1 2 base = m a s a s /2 m 1 m ( height) big 1 ( 2 base) ( height) small # 5 m ( 1 2 # 2.5 m W = m adx = 3kg = 5.625J or = to = 5.m: 5m adx = + + # 5 m = m # 5 m W = m adx = 3kg 2 = 6J B) or = to = 5.m: 5m adx = + # 5 m = 3m # 5 m W = m a dx = 3kg 17.5 = 52.5J ( = # 5 m = 17.5 = 2 Use the work-energy theorem W = K = 1 2 mv 2 f 1 2 mv 2 i, v i = at x =. The speed at x = 5m is W = 52.5J = 1 2 mv 2 f v f = J 3kg = 5.9 m s
5 Angular Momentum Problem 6 In diagram below a 2kg rock is at point P traveling horizontally with a speed of 12 m/s. At this instant what is the magnitude and direction of the angular momentum? If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of the angular momentum? Direction perpendicular to x-y plane indicates +z out of the page indicates z into page Also +x right +y up Angular Momentum L = r p = m r v valid for point particle w.r.t. point O r, r = 8m Using the right hand rule on L = r p = m r v it is easy to see that the direction of L is z or into the page v or the magnitude L = L = mvr sin143.1 = 2kg 12m / s 8m.6 L= 115.2kg-m 2 /s. Torque due to gravity on particle w.r.t. point O. = r g, g = mg = 2kg 9.8m / = 19.6N r, r = 8m Using the right hand rule on = r g it is easy to see that the g direction of is +z or out the page. = r g sin53.1 = 8m 19.6N.8 = 125.4Nim This can be expressed in a different unit = 125.4kg m 2 / Using second law for rotation in terms of angular momentum = d L. Hence the net torque is the rate of change of angular momentum Since the rate of change of angular momentum = d L / has opposite direction (+z) compared to the direction of the current angular momentum L = r p = m r v (-z), the angular momentum is decreasing. Can you see the similarity with our much earlier discussion on linear kinetics? INAL COMMENT AND ADVICE: Angular momentum L and Torque depend on the origin (O). or example, L = and = may be zero for origin O, but nonzero L and in another origin O /. Study angular momentum problems and static equilibrium problems.
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