Oregon State University PH 211 Fall Term 2018

Prep 6-7Recommended finish date: Wednesday, November 7

The formats (type, length, scope) of these Prep problems have been purposely created to closely parallel those of a typical exam (indeed, these problems were taken from past exams). To get an idea of how best to approach various problem types (there are three basic types), refer to these Sample Problems.

2

1. Evaluate the following statements (T/F/N). As always, explain your reasoning.

a. A 30-kg child standing on the earth exerts a gravitational force on the earth that is twice as much as that of a 15-kg child standing next to her. True. The FG acting on each—therefore by each (this is Newton’s Third Law)—is simply mchildg. So twice the mchild would give twice the force magnitude.

b. If one object is accelerating, there must be another object that’s accelerating. True. If a net force has been exerted on mass 1 (since it’s accelerating) it must have been applied by another object (mass 2): Fapp.21 But by Newton’s Third Law, a force was applied on mass 2 —equally but oppositely: Fapp.12 = –Fapp.21. So mass 2 must be accelerating, also.

c. Units of acceleration could be correctly written also as N/kg. True. The units of force, F, must be equivalent to m·a, so a is equivalent to F/m, which are N/kg in the SI system.

d. A 6-kg block on earth would be a 1-kg block on the moon. False. The mass of an object is not affected by which planet it’s near. Mass in intrinsic to the amount of matter contained in an object. If that hasn’t changed, the mass hasn’t changed.

e. A body that has zero acceleration may have forces acting on it. True. Zero acceleration indicates only that the (vector) sum of all forces acting on the object is zero. That sum may be (and is often) comprised of two or more force vectors.

f. A car is waiting at rest at a stop light on a westbound road. When the light turns green and the car accelerates forward (to the west), its rear bumper exerts a force on the rest of the vehicle, and that force is directed to the east.

True. Analyze the bumper and the rest of the car as two connected masses accelerating together. If the bumper is accelerating to the west, the rest of the car must be pulling on the bumper in that direction. So, by Newton’s 3rd Law, the bumper is pulling oppositely (east) on the rest of the car.

g. An object (m = 5.00 kg), initially at rest, experiences a net force of [1.5t2, –2t] N (where t is in seconds) throughouttheinterval0≤t≤4.00s.Thenallforceceases.Att = 7.50 s, the object’s speed is approximately5.54m/s.

False.

Pf = Pi + DP, but Pi = 0 (object starts at rest), so Pf = DP. Thus: mvf = J So: |vf| = |J|/m

J = ∫ F(t) dt = [∫ 1.5t2 dt , ∫ –2t dt] = [(1/2)t3|0

4 , –t2|

0

4] = [32 , –16] kg·m/s

Therefore: |J| = √[322 + (–16)2] √(1280) So: |vf| = |J|/m = √(1280)/5.00 = 7.16 m/s

<--- a

(car) (rear bumper)

3

2. For each item, be sure to show your work and/or explain your reasoning.

a. A 150,000 kg jet airliner is moving with constant velocity at an angle of 12.7° above the horizontal, at an altitude of 10 km and a speed of 200 m/s. What is magnitude of the total force acting on the airplane? Zero. By Newton’s First Law, if the object’s acceleration is zero (true here), so is the net force acting on it.

b. A 2-kg object has the acceleration vector [–1, 6] m/s2. What additional force must be applied to the object to put it into equilibrium? Express this as a magnitude and direction (using the +x-axis as 0°). Right now the net force on the object is SF = ma = 2[–1, 6] = [–2, 12] N Equilibrium is when SF = 0. To make this true, we must add the opposite of the current net force. Thus, we must add Fadd = [2, –12] N = √(22 + (–12)2)∠[tan-1(–12/2)] = 12.2 N∠–80.5°

c. A 2-kg object is constrained to moving along a straight line. The velocity as a function of time is shown here. (i) What is the net force acting on the object at t = 5s? Explain your reasoning. Fnet = SF = ma. a(5) is the slope of the v-t graph at t = 5: So a(5) = [v(5) – v(1)]/(5 – 1) = – 5 m/s2. Thus Fnet(5) = 2(–5) = –10 N (ii) What is the average net force acting on the object between t = 0s and t = 7s? Explain your reasoning. (Fnet)avg = maavg = m(vf – vi)/Dt = 2(–10 – 0)/7 = –2.9 N

d. Boxes A and B are sliding to the right across a frictionless table. The hand, H, is slowing them down. The mass of A is larger than the mass of B. Rank, from largest to smallest, these horizontal force magnitudes: FAB, FAH, FBH, FBA, FHA, FHB First, note: The hand isn’t touching A, so FAH = FHA = 0. Then note: The hand must exert enough force to accelerate the combined mass, but B must exert on A only enough force to accelerate the mass of A: FBA < FHB = 0. Final ranking (using Third Law equivalencies): FBH = FHB > FBA = FAB > FAH = FHA = 0.

4

3. For each item, be sure to show your work and/or explain your reasoning.

a. Thefigureshowstwomassesatrest.Thestringismassless,andthepulleysare frictionless. The spring scale reads in kg. What is the reading of the scale? Be sure to fully explain your reasoning. The reading is 5 kg. A scale reads only the force it is exerting in one direction. If it were to add or subtract the forces it is exerting at both ends (yes, a FBD will show this), it would read either double or zero, and that would make it useless.

b. In the situation shown, the horizontal surface is frictionless, and the pulley is massless and frictionless. Is the tension in the cable less than, greater than, or equal to 88.2 N? Do a full analysis with a FBD to support your argument. The hanging mass (m1): The sliding mass (m2):

The hanging mass (m1): The sliding mass (m2): But note: SF1.y = m1a1.y SF2.x = m2a2.x FT.12 = FT.21 (These are magnitudes and so are equal.) FG.P1 – FT.21 = m1a1.y FT.12 = m2a2.x a2.x = a1.y (Their magnitudes and signs are equal.) m1g – FT.21 = m1a1.y FT.12 = m2a2.x So call the above simply FT and a. Thus: m1g – FT = m1a and: FT = m2a Solve one equation for a: a = FT/m2 Substitute and solve for FT: m1g – FT = m1FT/m2 FT = m1g/(1 + m1/m2) = (9)(9.80)/(1 + 9/5) = 31.5 N (less than 88.2 N)

c. Rank the tension in the string, from greatest to least, in the cases shown here. Assume in each case that the hanging mass is the same, that friction is negligible, and that the pulley and rope are both massless. Use FBDs! The solution for FT here is the same as in 2B, above: FT = m1g/(1 + m1/m2) Here m1 is the same in all three cases, but not m2. And when m2 is larger the denominator here is smaller. That is, when m2 is larger, FT is larger. Since m2A > m2B > m2C, this means that FTA > FTB > FTC.

d. An object of mass M is held in place by an applied force F and a pulley system, asshowninthefigure.Thepulleysaremasslessandfrictionless,asarethecables. a. Find the tension in rope section T1. FT.1 = F = Mg/2 (See below for reasoning.) b. Find the tension in rope section T2. FT.2 = F = Mg/2 (See below for reasoning.) c. Find the tension in rope section T3. FT.3 = F = Mg/2 (See below for reasoning.) d. FindthetensioninropesectionT4. FT.4 = 3Mg/2 (See below for reasoning.) e. Find the tension in rope section T5. FT.5 = Mg (FT.5 holds M at rest against FG.PM.) f. Find the magnitude of F. F = Mg/2 (FT.2 + FT.3 = FT.5. So: F + F = Mg) For any massless cable, the tension at any point in it is the same (which you can prove by using a FBD to analyze any two parts of it). And if such cables are connected to ideal (massless, frictionless) pulleys, their tension is redirected without alteration. And looking at the whole pulley contraption as one object, we have just one upward force (FT.4) and two downward forces (F and Mg) acting on it. Thus: FT.4 = F + Mg.

5 kg

9 kg

y

x

FG.P1

FT.21

FG.P2

FN.2

FT.12

M

F

T1

T2T3

T4

T5

5

3. e. For the situation shown, where four ideal (massless) pulleys support two masses hanging at rest, evaluate (T/F/N) the following statement. As always, you must justify your answer with a valid mix of words, drawings and calculations.

FTB/FTA = 8

False. Each ideal pulley transmits the tension force throughout any continuous wire. (This is how we know that the tension in the wire connecting the second pulley to the ceiling is also FTA.) Thus, each pulley experiences twin upward forces; so the tension each supports downward is twice that. Hence the labels shown here. Thus: FTB/FTA = 4

m2

FTA

FTB

m1

m2

FTA

FTB

m1

FTA

2FTA

4FTA

6

4. a. Evaluate the following statements (T/F/N). As always, explain your reasoning. (i) If an elevator is moving vertically upward, then the weight of a person standing in that elevator is always greater than her weight when she is at rest. False. If she is slowing down or moving at constant speed, her weight is either less than or equal to her rest weight. (ii) The weight of an astronaut orbiting the earth is greater than his weight when orbiting the moon. False. Not greater than—equal to. The astronaut is weightless in either case: 0 = 0. (iii)Thestaticfrictionforceexertedbyonesurfaceonanotherdependsonlyonthefrictioncoefficient(mS) and the normal force that exist between the two surfaces. False. The static friction force varies (from 0 to FS

max) in response to the applied force(s) that would other- wise be causing the surfaces to slide past each other. Only FS

max is calculated via mSFN. (iv) A stationary object can exert a kinetic friction force. (Any answer OK here). As some students have alertly pointed out, there’s no object in the universe that is known to be stationary. So by that assumption, this statement has no meaning—it’s either false or not enough information. However, if we designate something such as the earth as “stationary,” then certainly it can exert a kinetic friction force: Slide a book across the floor and watch it slow down and stop. (v) A static friction force can cause acceleration. True. For example, look at the force FS.12 in exercise 9 of Lab 5-I (the FBD exercise set). (vi)Whenablockslidesat4m/sonasurfacewithfriction,thereistwicethefrictionforceactingonitas when it slides on that surface at 2 m/s. False. No, for low speeds, mK (and therefore FK) is about equal for any non-zero speed (so long as FN has not changed). (vii) A friction force between two surfaces acts on each surface equally in magnitude but opposite in direction. True. This is just a re-statement of Newton’s Third Law: Any force acting between two objects acts on each with the same magnitude but opposite direction. (viii)Thecoefficientofkineticfrictionisaforce. False. mK is a unitless ratio of two force magnitudes: mK = FK.S1/FN.S1 (ix)Anormalforceisdefinedasaforceexertedverticallyupwardbyasurface. False. A normal force is defined as a force exerted by a surface in a direction perpendicular to that surface. That direction is not necessarily vertical or upward. (x) Amanstandingonthelevelfloorofanelevatorthatismovingverticallydownwardcouldhaveaweight greater than the gravitational force acting on him. True. The man (and elevator) could be accelerating upward if the elevator were slowing while moving downward. And an upward acceleration would mean that the normal force upward by the elevator floor (what the man feels to judge his weight) must be must be greater than the downward gravitational force. (xi) The gravitational force exerted by the earth on the orbiting space shuttle is greater than the gravitational force exerted by the earth on the pilot’s seat inside that shuttle. True. |FG.P1| = m1g, where g is the local value of gravitational free-fall acceleration. And the same local g value applies to the entire space shuttle and to all of its parts, so a larger mass experiences a larger gravitational force acting on it.

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4. b. Helenispushingaboxsothatitslidesacrossthefloor.Thespeedoftheboxisincreasing. Evaluate the following statements (T/F/N). As always, explain your reasoning. (i) The frictional force is negligible (or effectively zero.) Not enough information. All we know from the given information is that the box is speeding up, so the horizontal component of her pushing force exceeds the kinetic friction force opposing the sliding. That is: FApp.HBcosq > FK, where q is the angle of her push, as measured from the horizontal. In other words: FApp.HBcosq > FK. But we simply don’t know how much FK is. (ii) The frictional force is balanced by Helen’s applied force. False. The box is accelerating, so the friction force is not being balanced; it is being exceeded. (iii) Helen is pushing with a force greater than the normal force on the box. Not enough information. As argued in (i), above, FApp.HBcosq > mKFN. And FN = mg – FApp.HBsinq. Thus, FApp.HBcosq > mK(mg – FApp.HBsinq). But we don’t know the value of either mK, m or q. (iv) Helen’s pushing force on the box is greater than the frictional force on the box. True. As argued in (i), above, FApp.HBcosq > FK, and so even if Helen’s entire force is directed horizontally (q = 0), that would still mean FApp.HB > FK. (v) The box’s weight is less than Helen’s applied pushing force. Not enough information. The box’s weight is the vector sum of all contact forces (FApp.HB, FN and FK) being exerted on it. And we don’t know the magnitudes of any of those forces (nor the direction of FApp.HB).

8

5. a. The block (m = 30 kg) is on a level surface with friction coefficients ms = 0.750 and mk = 0.500. If FT1 = 100 N, q1 = 40.0°, and q2 = 25.0°, find FT2 so that:

– the block is still at rest but “just about” to start sliding to the right.

– the block slides to the right at constant velocity.

m

FT1

q1

FT2

q2

FBD of m—situation (a)

coord. axes

FG

FN

Fmaxs

q2q1

FT2

FT1

Fmaxs becomes Fk

in situation (b).

y

xx-analysis—situation (a)

SFx = max

FT2x – FT1x – Fmaxs = max

FT2cosq2 – FT1cosq1 – msFN = 0

y-analysis—situation (a)

SFy = may

FN + FT1y + FT2y – FG = may

FN + FT1sinq1 + FT2sinq2 – mg = 0

x-analysis—situation (b)

SFx = max

FT2x – FT1x – Fk = max

FT2cosq2 – FT1cosq1 – mkFN = 0

y-analysis—situation (b)

SFy = may

FN + FT1y + FT2y – FG = may

FN + FT1sinq1 + FT2sinq2 – mg = 0

These are two equations in two unknowns (FN and FT2)—solve simultaneously:

Solve for FN: FN = mg – FT1sinq1 – FT2sinq2

Substitute: FT2cosq2 – FT1cosq1 – ms(mg – FT1sinq1 – FT2sinq2) = 0

Collect terms: FT2cosq2 + msFT2sinq2 = FT1cosq1 + msmg – msFT1sinq1

Solve: FT2 = (FT1cosq1 + msmg – msFT1sinq1)/(cosq2 + mssinq2)

= [(100)(cos40) + (.75)(30)(9.80) – (.75)(100)(sin40)]/[cos25 + (.75)(sin25)] = 203 N

For (b): FT2 = (FT1cosq1 + mkmg – mkFT1sinq1)/(cosq2 + mksinq2)

= [(100)(cos40) + (.50)(30)(9.80) – (.50)(100)(sin40)]/[cos25 + (.50)(sin25)] = 171 N

9

5. b. A 3-kg block is free to slide vertically up or down a frictionless wall. A 20 N force is applied as shown. Find the block’s acceleration (both magnitude and direction). SFx = max SFy = may FN – Fx = max Fy – FG = may FN – Fsin60° = 0 Fcos60° – mg = may (Fcos60°)/m – g = ay ay = –6.47m/s2 (that’s 6.47 m/s2 downward)

c. A 3-kg block is held at rest against a wall by a two identical applied forces F,asshown.Thewallhasacoefficientofstatic friction of 0.65 with the block. Find the magnitude of F at which the block would be ready to slip down the wall.

SFx = max SFy = may FN – F = max F + Fs

max – FG = may FN – F = 0 F + msFN – mg = 0 FN = F F + msF – mg = 0 F + msF – mg = 0 F(1+ ms) = mg F = mg/(1+ ms) = 17.8 N

d. A box of mass m is initially sliding at a speed of v on a horizontal surface. Wind is applying a constant force magnitude F at an angle q down from the horizontal, in the opposite direction of the box’s motion. The box comes to rest after traveling a distance d.Findthecoefficientofkineticfriction,mK, between the box and the surface, expressed in terms of m, v, F, q , d and g. First, do kinematics to find the x-acceleration value necessary to bring the box to rest in distance d: 02 = v2 + 2axd Thus: ax = –v2/(2d)

Now do a FBD analysis (referring to diagrams here—using conventional x- and y- axes): SFx = max SFy = may –FK – Fx = max FN – Fy – FG = may mKFN – Fcosq = –mv2/(2d) FN – Fsinq – mg = 0

Solve y-equation for FN: FN = Fsinq + mg Substitute into x-equation: mK(Fsinq + mg) – Fcosq = –mv2/(2d) Solve for mK: mK = [Fcosq – mv2/(2d)]/(Fsinq + mg)

FF

60°20 N

y

x

FG

F

FN

y

x

FG

F

FNF

FSmax

Fq

FG

F

FN

q

FK

10

5. e. Two identical masses are connected by a massless cord, as shown. They are being pulled to the right on a level, frictionless surface by horizontal force F. The tension in the cord, the angle q of the cord, and F all remain constant. Evaluate (T/F/N) the statement below. As always, you must justify your answers with a valid mix of words, drawings and calculations.

The difference in magnitude between the two normal forces being exerted in this situation is F·tanq.

True.

m1 m2

SF1.x = m1a1.x SF1.y = m1a1.y SF2.x = m2a2.x SF2.y = m2a2.y

FT.21.x = m2a1.x FN.1 + FT.21.y – FG.1 = m1a1.y F – FT.12.x = m2a2.x FN.2 – FT.12.y – FG.2 = m2a2.y

FTcosq = ma FN.1 + FTsinq – mg = 0 F – FTcosq = ma FN.2 – FTsinq – mg = 0 FN.1 = mg – FTsinq FN.2 = mg + FTsinq

So: FTcosq = F – FTcosq

Or: FT = F/(2cosq) Thus: FN.2 – FN.1 = (mg + FTsinq) – (mg – FTsinq) = 2FTsinq

= 2sinq[F/(2cosq)] = F·tanq

Fq

FG.1

FT.21

q

FN.1

FG.2FT.12

q

FN.2

F

11

6. In each case below, the acceleration magnitude on the 5-kg box (M)is4m/s2, which is as large as possible with-outtheboxsliding.Foreachcase,findthecoefficientofstaticfrictionbetweentheboxandthesurfacebeneathit.

a. The box is on the bed of a truck that is accelerating on level ground. SFx = max SFy = may Fs

max = max FN – FG = may msFN = max FN – mg = 0 Therefore: msmg = max Thus: ms = ax/g = 4/9.80 = 0.408

b. The box is in a ski gondola on its way to the top of the mountain. (The gondola does not swing during the motion; its attachment to its cable remains vertical at all times.) SFx = max SFy = may Fs

max = max FN – FG = may msFN = m(acos30°) FN – mg = m(asin30°) Therefore: ms(masin30° + mg) = m(acos30°) Thus: ms = (acos30°)/(asin30° + g) = (4cos30°)/(4sin30° + 9.80) = 0.294 FG

FN

FSmax

FG

FN

FSmax

12

7. a. Block 2 is hanging freely (at rest) by a thread from block 1. Block 1 is not attached to the ceiling, but it is being held at rest against the ceiling, due to the effects of the force, F. Find the maximum force magnitude, F, which can be applied so that block 1 won’t slip. q = 25.0°; m1=4.06kg;m2 = 1.37 kg mS.ceiling.block1 = 0.890

For m2: SFx = m2ax For m1: SFx = m1ax 0 = 0 Fx – FS

max = m1ax Fsinq – mSFN = 0

SFy = m2ay SFy = m1ay FT1.2 – FG.2 = m2ay Fy – FT2.1 – FG.1 – FN = m1ay FT1.2 – m2g = 0 Fcosq – FT2.1 – m1g – FN = 0

Thus: FT1.2 = m2g = 1.37(9.80) = 13.426 N Also: FN = Fsinq/mS So: FT2.1 = FT1.2 = 13.426 N So: Fcosq – FT2.1 – m1g – Fsinq/mS = 0

Solve for F: Fcosq – Fsinq/mS = FT2.1 + m1g F(cosq – sinq/mS) = FT2.1 + m1g F = (FT2.1 + m1g)/(cosq – sinq/mS) = [13.426 + (4.06)(9.80)]/[cos25° – (sin25°)/0.890] = 123 N

b. Block 2 is attached to the side of block 1 only by a thread. The blocks are accelerating together as Block 1 slides along a horizontal shelf. Findthecoefficientofkineticfriction,mK. q = 25.0°; F = 98.0 N; m1=4.65kg;m2 = 1.73 kg

For m2: SFx = m2ax For m1: SFx = m1ax FT1.2.x = m2ax Fx – FK – FT2.1.x = m1ax FT1.2sinq = m2ax Fcosq – mKFN – FT2.1sinq = m1ax

SFy = m2ay SFy = m1ay FT1.2.y – FG.2 = m2ay FN – FT2.1.y – FG.1 – Fy = m1ay FT1.2cosq – m2g = 0 FN – FT2.1cosq – m1g – Fsinq = 0

FT1.2 = m2g/cosq = 1.73(9.80)/cos25° = 18.707 N FN = FT2.1cosq + m1g + Fsinq FT2.1 = FT1.2 = 18.707 N = 18.707cos25° + 4.65(9.80) + 98sin25° ax = FT1.2sinq/m2 = 18.707sin25°/1.73 = 4.5699 m/s2 = 103.94 N

Solve for mK: mK = (Fcosq – FT2.1sinq – m1ax)/FN = [98cos25° – 18.707sin25° – 4.65(4.5699)]/103.94 = 0.574

m1

m2

mK

Fq

q

q

m1

m2

mS

F

FG.2

FT1.2

qF

FG.1

FT2.1 FN

FSmax

FG.2

FT1.2q

q

FG.1

FT2.1

FN

FK

q

13

8. a. A block (m = 10.2 kg) sits at rest on a surface that has µS and µK values (with the block of 0.75 and 0.36, respectively. The surface is initially level but then is gradually tilted until the block slides down the slope. At that angle of tilt, how long does it take the block to slidefor1.84malongthatslope—measuredfromthe moment it begins to move? (Ignore any sudden jerks or irregularmotiontobeginwith—treatthemotionasa smooth transition from rest.)

The first situation to analyze is the block while it’s still at rest but about to slip:

The unknowns here are FN and q. And it’s really the q value we want to determine—the slope angle at which the block will slide down the incline—so solve for FN first (using the y-equation) and substitute this into the x-equation:

FN = mgcosq (from the y-equation)

mgsinq – mS(mgcosq) = 0 (substituting into the x-equation)

mgsinq = mS(mgcosq) (rearranging)

sinq = mScosq (simplifying)

tanq = mS (simplifying)

q = tan-1(mS) = tan-1(0.75) = 36.8699° This is the slope at which the block will be sliding.

q

m

FBD

coord. axes

x-analysis

SFx = max

FGx – FSMax = max

mgsinq – mSFN = 0

y-analysis

SFy = may

FN – FGy = may

FN – mgcosq = 0

y

x

q FG

FN

FSMax

14

The next situation to analyze is the block while it’s sliding down the slope:

The unknowns here are FN and ax. And it’s really the ax value we want to determine—the block’s acceleration—so solve for FN first (using the y-equation) and substitute this into the x-equation:

FN = mgcosq (from the y-equation)

mgsinq – mK(mgcosq) = max (substituting into the x-equation)

mg(sinq – mKcosq) = max (simplifying)

g(sinq – mKcosq) = ax = (9.80)[sin(36.8699) – .36·cos(36.8699)] = 3.0576 m/s2

Now it’s just a kinematics problem. Knowing Dx = 1.84, v0 = 0, and a = 3.0576, we can solve for Dt:

Dx = v0(Dt) + (1/2)a(Dt)2

Dx = (1/2)a(Dt)2 (simplifying)

2Dx/a = (Dt)2 (rearranging)

[2Dx/a]1/2 = Dt = [2(1.84)/3.0576]1/2 = 1.10 s

q

m

FBD

coord. axes

x-analysis

SFx = max

FGx – FK = max

mgsinq – mKFN = max

y-analysis

SFy = may

FN – FGy = may

FN – mgcosq = 0

y

x

q FG

FN

FK

15

8. b. The block shown at right remains at rest while the force F is present. Find the block’s speed 2.50 seconds after F is suddenly removed. Be sure to show your work and/or explain your reasoning. Does the block move when released? Is the slope sufficient to cause it to slip? Ask that question first! If the gravity force down the slope exceeds the maximum static friction force the surface can exert, it will slip. SFx = ? SFy = may

Fsmax – FG.x = ? FN – FG.y = may

ms FN – mgsinq = ? FN – mgcosq = 0

FN = mgcosq = 10(9.80)cos20° = 92.09 So: Fsmax = ms FN = .50(92.09) = 46.04 N

mgsinq = 10(9.80)sin20° = 33.52 N Since the gravitational pull down the slope is less than Fs

max, the block will not slip; its speed will remain zero.

c. A 3-kg block is sliding at constant speed down aslopedsurface,asshown.Thecoefficientof kinetic friction (mk) between the block and the slopeis0.45.Whatistheangleq of the slope? Show your work and/or explain your reasoning. SFx = max SFy = may FG.x – Fk = max FN – FG.y = may mgsinq – mkFN = 0 FN – mgcosq = 0 FN = mgcosq

mk = mgsinq/FN = mgsinq/(mgcosq) = tanq Therefore: q = tan-1(mk) = 24.2°

d. A 3-kg block is sitting at rest on a sloped surface, as shown. Thecoefficientofstaticfriction(ms) between the block and the sloped surface is 0.55. What is the magnitude of the total force exerted by the sloped surface on the block? Be sure to show your work and/or explain your reasoning. SFx = max SFy = may FG.x – FS = max FN – FG.y = may mgsinq – FS = 0 FN – mgcosq = 0 FS = mgsinq = 10.055 N FN = mgcosq = 27.627 N

The total force by slope on block is a vector sum: Fslope.block = FN + FS

But FN and FS are perpendicular to each other, so they form a right triangle whose hypotenuse is the total force; its magnitude can be calculated via the Pythagorean Theorem: 29.4N

Also OK is clearly explaining: The only object supporting the block upward is the sloped surface, so the sum of all forces exerted by the surface on the block must exactly counter the downward gravitational force on the block (FG = mg = 3·9.80 = 29.4 N).

20°

q°

F

coord. axes

y

x

FG

FN

Fs

q

10 kg

20°

mk = 0.25

ms = 0.50

y

x

FG

FN

Fk

q

y

x

FG

FN

Fs

q

FN

Fs

Fslope.block

q

16

9. For each item, be sure to show your work and/or explain your reasoning.

a. A200ghockeypuckislaunchedupametalrampthatisinclinedata30°angle.Thecoefficientsofstatic andkineticfrictionbetweenthehockeypuckandthemetalrampareμS=0.40andμK = 0.30, respectively. The puck’s initial speed is 3 m/s. (i) Find the net force magnitude and direction on the puck while it is sliding up the ramp. SF1x = m1a1x –FG.1.x – FK.1 = m1a1x –m1gsinq – mK.1FN.1 = –m1a

SF1y = m1a1y FN.1 – FG.1.y = m1a1y FN.1 – m1gcosq = 0

So: FN.1 = m1gcosq Thus: –m1gsinq – mK.1(m1gcosq) = –m1a Find a: a = g(sinq + mK.1cosq) = (9.80)[sin30° + (0.30)cos30°] = 7.446 m/s2

This is the total acceleration magnitude of the puck (since ay = 0), and it’s in the negative x-direction, so a = ax = –7.446 m/s2

SF1 = m1a = (0.200)(–7.446) = 1.49Ndowntheramp (ii) How far up the ramp does the puck travel? Just do kinematics, since the x-acceleration value is now known: vf.x

2 = vi.x2 + 2axDx Thus: Dx = (vf.x

2 – vi.x2)/(2ax) = (02 – 32)/[2(–7.446)] = 0.604m

b. A 20 kg block on a table is connected by a string to a 12 kg mass, which is hanging over the edge of the table (modeled here as hanging over an ideal pulley). The 20-kg blockis3.1mfromthetableedge.Thecoefficientofkineticfrictionbetweentheblock andthetableis0.1.Thecoefficientofstaticfrictionbetweentheblockandthetableis0.5. (i) Is the hanging block is heavy enough to cause the block on the table to slide? Could FS.1

max hold m1 against the tension of the string, FT (= FT.21 = FT.12), if m2 were hanging from rest? Find out: In that special, simple case: FN.1 = FG.P1 = m1g And: FT.12 = FG.P2 = m2g Thus: FS.1

max = mS(FN.1) = mS(m1g) = (0.5)(20)(9.80) = 98.0 N But: FT.21 = FT.12 = m2g = (12)(9.80) = 118 N The tension FT.21 would be greater than FS.1

max could hold. The block will slip. (ii) Regardless of your answer to part a, assume that the block on the table does slide and determine how much time will pass before the block reaches the end of the table. The sliding mass (m1): The hanging mass (m2): SF1.x = m1a1.x SF1.y = m1a1.y

FT.21 – FK.1 = m1a1.x FG.P2 – FT.12 = m2a2.y FT – FK.1 = m1a m2g – FT = m2a FT = FK.1 + m1a FT = m2g – m2a So: FK.1 + m1a = m2g – m2a Or: a = (m2g – FK.1)/(m1 + m2) = (m2g – mKm1g)/(m1 + m2) = [(12)(9.80) – (0.1)(20)(9.80)]/(20 + 12) = 3.063 m/s2

Now kinematics: Dx = vi.x(Dt) + (1/2)ax(Dt)2 Since vi.x = 0, this simplifies to Dx = (1/2)ax(Dt)2

Thus: Dt = √[2(Dx)/ax] = √[(2)(3.1)/3.063] = 1.42s

30°y

x

FG

FN

FK

q

20 kg

12 kgy

x

FG.P1

FN.1

FT.21

FS.1max

FG.P2

FT.12

FG.P1

FN.1

FT.21

FK.1

FG.P2

FT.12

17

c. Thefigureshowsablockofmassm resting on a 20° slope. The block and thissurfacehaveacoefficientofstaticfrictionof0.86andacoefficientof kinetic friction of 0.76. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg. (i) What is the minimum mass m that will stick and not slip?

The sticking mass (m1): The hanging mass (m2):

SF1.x = m1a1.x SF2.x = m2a2.x

FT.21 – FG.P1.x – FS.1max = m1a1.x 0 = 0

FT – m1gsinq – mSFN.1 = 0

SF1.y = m1a1.y SF2.y = m2a2.y

FN.1 – FG.P1.y = m1a1.y FT.12 – FG.P2 = m2a2.y FN.1 – m1gcosq = 0 FT – m2g = 0

FN.1 = m1gcosq FT = m2g

So: m2g – m1gsinq – mSm1gcosq = 0 Or: m1 = m2/(sinq + mScosq) = 2/[sin20° + (0.86)(cos20°)] = 1.7389 = 1.74kg

(ii) Ifamasshalfofthatvaluewereplacedthereinstead,findtheaccelerationoftheblocks.

Sliding mass (m1 = 1.7389/2 = 0.8694): Hanging mass (m2):

SF1.x = m1a1.x SF2.x = m2a2.x

FT.21 – FG.P1.x – FS.1max = m1a1.x 0 = 0

FT – m1gsinq – mKFN.1 = m1a

SF1.y = m1a1.y SF2.y = m2a2.y

FN.1 – FG.P1.y = m1a1.y FT.12 – FG.P2 = m2a2.y FN.1 – m1gcosq = 0 FT – m2g = –m2a

FN.1 = m1gcosq FT = m2g – m2a

So: m2g – m2a – m1gsinq – mKm1gcosq = m1a Or: m2g – m1gsinq – mKm1gcosq = a(m1 + m2) Thus: a = [m2g – m1gsinq – mKm1gcosq]/(m1 + m2) = [(2)(9.80) – (0.8694)(9.80)(sin20°) – (0.76)(0.8694)(9.80)(cos20°)]/(0.8694 + 2) = 3.69 m/s2

!y

x

FG.P1

FN.1

FS.1max

q

FG.P2

FT.12y

x

y

x

FG.P2

FT.12y

x

FT.21

FG.P1

FN.1

FK.1

q

FT.21

18

9. d. A box (m = 10.0 kg) on a level surface has an applied pushing force, F, being exerted on it at an angle, q, as shown.Thestaticandkineticfrictioncoefficientsbetweenboxandsurfaceare0.700and0.350,respectively.

a. Find the friction force magnitude acting on the box bottom when F = 75.0 N and q = 5.00°.

b. Find the friction force magnitude acting on the box bottom when F = 75.0 N and q = 20.0°.

c. What minimum value of F would be required to move the box when q=40.0°?

d. What is the maximum angle(0≤q≤90°)atwhichyoucouldexerttheforceF (of any magnitude) toward the box and still get the box to move?

Here’s a situation where there are two different values (F and q) that you can vary. For any given value of F, if you want it to be the force Fslip that puts the block into “ready-to-slip” mode, there’s just one certain angle (if any) that you could apply that force (a certain value of q)—call this qslip. So how do Fslip and qslip relate to each other? Find out: Analyze the situation as if the box (call it m1) were about to slip:

SF1.x = m1a1.x SF1.y = m1a1.y FS.1

max – Fx = m1a1.x FN.1 – Fy – FG.1 = m1a1.y mSFN.1 – Fslipcosqslip = 0 FN.1 – Fslipsinqslip – m1g = 0

Thus: FN.1 = m1g + Fslipsinqslip

Thus: mS(m1g + Fslipsinqslip) – Fcosqslip = 0

Or: Fslipcosqslip – mSFslipsinqslip = mSm1g

So: Fslip = mSm1g/(cosqslip – mSsinqslip)

a. Find Fslip for qslip = 5.00°: Fslip = (0.7)(10)(9.80)/[cos(5°) – (0.7)sin(5°)] = 73.4 N

Since the actual F applied (75 N) is greater than Fslip for this q, the block slips; the friction force is kinetic. But the above y-analysis doesn’t change for the sliding case, so: FN.1 = m1g + Fsinq

Thus: FK.1 = mK(FN.1) = mK(m1g + Fsinq)

= (0.350)[(10)(9.80) + (75.0)sin(5.00°)] = 36.6 N

b. Find Fslip for qslip = 20.0°: Fslip = (0.7)(10)(9.80)/[cos(20°) – (0.7)sin(20°)] = 98.0 N

Since the actual F applied (75.0 N) is less than Fslip for this angle, the block does not slip, so the friction force is static but not maximum. So the analysis becomes:

SF1.x = m1a1.x FS – Fx = m1a1.x FS – Fcosq = 0

Thus: FS = Fcosq

= (75.0)cos20° = 70.5 N

c. Find Fslip for qslip = 40.0°: Fslip = (0.7)(10)(9.80)/[cos(40°) – (0.7)sin(40°)] = 217 N

d. Looking at Fslip = mSm1g/(cosqslip – mSsinqslip), notice that Fslip approaches infinity when the denominator (cosqslip – mSsinqslip) approaches zero. In other words when (cosqslip – mSsinqslip) goes to zero, it doesn’t matter how great the force you apply at that qslip, the block will still be only “about to slip”—still at rest.

So the maximum angle, qslip.max, is given by: cosqslip.max – mSsinqslip.max = 0

That is: tanqslip.max = 1/mS Or: qslip.max = tan-1(1/mS) = tan-1(1/0.7) = 55.0°

Fq

y

xFG.1

FS.1max

q

FN.1

F

y

xFG.1

FS

q

FN.1

F

19

10. a. Thefigureshowsa100kgblockbeingreleasedfromrestfromaheightof1.0m. Theblocktakes0.64storeachthefloor.Whatisthemassoftheblockontheleft? First, do kinematics to find the y-acceleration value so that the box lands in 0.64 s: Dy = vi.y(Dt) + (1/2)ay(Dt)2 Or: ay = 2[Dy – vi.y(Dt)]/(Dt)2 = 2[–1 – 0(0.64)]/(0.64)2 = –4.883 m/s2

Now do a FBD analysis (referring to diagrams here—using conventional x- and y- axes): Let the unknown mass be m1 and the 100-kg mass be m2. And note that the magnitudes of their accelerations will be the same (a = 4.883 m/s2); and that the tension forces exerted will also be equal (because the pulley is ideal—massless). Thus: m1: m2: SF1y = m1a1y SF2y = m2a2y FT2.1 – FG.1 = m1a1y FT1.2 – FG.2 = m2a2y FT – m1g = m1a FT – m2g = –m2a

So: FT = m2g – m2a Then: m2g – m2a – m1g = m1a Or: m2(g – a) = m1(a + g) So: m1 = m2(g – a)/(a + g) = 100(9.80 – 4.883)/(4.883 + 9.80) = 33.5 kg

b. The two blocks shown here are sliding down the incline. Find the tension in the (massless) string. Do a FBD analysis (referring to diagrams here—using axes as shown): Let the 1-kg mass be m1 and the 2-kg mass be m2. Note that the magnitudes of their accelerations will be the same (a). Thus:

m1: m2:

SF1x = m1a1x SF2x = m2a2x FT2.1 + FG.1.x – FK.1 = m1a1x FG.2.x – FT1.2 – FK.2 = m2a2x FT + m1gsinq – mK.1FN.1 = m1a m2gsinq – FT – mK.2FN.2 = m2a

SF1y = m1a1y SF2y = m2a2y FN.1 – FG.1.y = m1a1y FN.2 – FG.2.y = m2a2y FN.1 – m1gcosq = 0 FN.2 – m2gcosq = 0

So: FN.1 = m1gcosq And: FN.2 = m2gcosq Thus: FT + m1gsinq – mK.1(m1gcosq) = m1a And: m2gsinq – FT – mK.2(m2gcosq) = m2a Find a: a = FT/m1 + gsinq – mK.1(gcosq) And: a = gsinq – FT/m2 – mK.2(gcosq) So: FT/m1 + gsinq – mK.1(gcosq) = gsinq – FT/m2 – mK.2(gcosq) Find FT: FT(1/m1 + 1/m2) = (mK.1 – mK.2)(gcosq) FT = [(mK.1 – mK.2)(gcosq)]/(1/m1 + 1/m2) = [(0.20 – 0.10)(9.80)(cos20°)]/(1/1.0 + 1/2.0) = 0.614N

!

FG.2

FT1.2

FG.1

FT2.1

y

xFG.1

FN.1

FK.1

q

FT.21

FG.2

FN.2

FK.2

q

FT.12

20

x-analysis

SFx = m1ax

FN.21 – Fk.1 = m1ax

FN.21 – mkFN.1 = m1ax

y-analysis

SFy = m1ay

FN.1 – Fs.21max – FG.1 = m1ay

FN.1 – msFN.21 – m1g = 0

FBD of m1

FBD of m2

x-analysis

SFx = m2ax

–FN.12 = m2ax

–FN.12 = m2ax

y-analysis

SFy = m2ay

Fs.12max – FG.2 = m2ay

msFN.12 – m2g = 0

coord. axes

y

x

FG.1

FN.1

FN.21

Fk.1

coord. axes

y

x

FG.2

Fs.21max

Fs.12max

FN.12

m2 m1

10. c. Two blocks (masses m1 = 26.5 kg; m2 = 38.9 kg) are moving together horizontally to the right. Block 2 is touching, but not attached to, block1.Thecoefficientofstaticfriction,ms, between the two blocks is1.47(yes,theyarequitesticky).Findtheminimumcoefficientof kinetic friction, mk,betweenblock1andthefloorsothatblock2 does not slip.

First, do a complete FBD and Newton’s Laws analysis of each mass, as follows:

21

Now solve for mk, as follows:

I. Solve msFN.12 – m2g = 0 for FN.12: FN.12 = m2g/ms II. Solve –FN.12 = m2ax for ax (the x-acceleration of both masses): ax = –FN.12/m2 = –(m2g/ms)/m2 = –(g/ms) III. Solve for FN.21: By Newton’s Third Law, the magnitudes of FN.21 and FN.12 must be equal. FN.21 = FN.12 = m2g/ms

IV. Solve FN.1 – msFN.21 – m1g = 0 for FN.1: FN.1 = msFN.21 + m1g = ms(m2g/ms) + m1g = (m1 + m2)g V. Solve FN.21 – mkFN.1 = m1ax for mk: mk = (FN.21 – m1ax)/FN.1 = [m2g/ms + m1(g/ms)]/[(m1 + m2)g] = 1/ms

10. d. As shown in this side (cutaway or “x-ray”) view, a sphere (msphere = 2.00 kg) rests in one compartment of a two-compartment box (mbox = 5.00 kg). The box rests on a level, frictionless surface. All surfaces of the box and sphere are also frictionless.

When the box and sphere are at rest (as shown), there are two normal forces acting on the sphere, and one of those forces has a magnitude 20% greater than the other.

But when a tension force, FT (not shown here) is then exerted horizontally (to the right) on the right side of the box, the two normal forces acting on the sphere become equal in magnitude. Find the magnitude of FT.

Let the two surfaces of the box that are exerting normal forces on the sphere be called “Wall” (W) and “Divider” (D). Then here is the analysis of the sphere (m1) when everything is at rest, as shown above (and note the definition of q now on that diagram):

SF1.x = m1a1.x SF1.y = m1a1.y FN.W1 – FN.D1.x = m1a1.x FN.D1.y – FG.1 = m1a1.y FN.W1 – FN.D1cosq = 0 FN.D1sinq – m1g = 0

Thus: FN.D1 = FN.W1/cosq

But we also know: FN.D1 = 1.20(FN.W1)

Therefore: 1.20(FN.W1) = FN.W1/cosq

Solve for q: q = cos-1(1/1.20) = 33.557°

Now analyze the combined mass (box and sphere together—call this m1&2), when the force FT is applied:

SF1&2.x = m1&2a1&2.x SF1&2.y = m1&2a1&2.y FT = m1&2a1&2.x FN.1&2 – FG.1&2 = m1&2a1&2.y FT = m1&2ax FN.1&2 – m1&2g = 0

Thus: ax = FT/m1&2

Now analyze just the sphere (m1) as it is accelerating to the right inside the box:

SF1.x = m1a1.x SF1.y = m1a1.y FN.W1 – FN.D1.x = m1a1.x FN.D1.y – FG.1 = m1a1.y FN.W1 – FN.D1cosq = m1ax FN.D1sinq – m1g = 0

Thus: FN.D1 = m1g/sinq

Thus: FN.D1 – FN.D1cosq = m1(FT/m1&2)

Or: (m1g/sinq)(1 – cosq) = m1(FT/m1&2)

Solve for FT: FT = (m1&2/m1)(m1g/sinq)(1 – cosq)

= (7.00/2.00)[(2.00)(9.80)/sin(33.557°)][1 – cos(33.557°)] = 20.7 N

FG.1

FN.W1q

FN.D1y

x

q

FG.1

FN.W1q

FN.D1y

x

y

x

FG.1&2

FT

FN.1&2

23

11. a. Two crates (m1 is known; m2 is unknown) are being dragged together (one linked to the other via a horizontal cable, as shown)acrossthefloorofalargefreightelevator.Thecrates’ horizontal velocity is constant. The elevator is accelerating upward at a known rate, a. The pulling force F is known, as is the angle q above the horizontal at which F pulls. Thecoefficientofkineticfriction,mK, between each crate andtheelevatorfloorisalsoknown.Findm2, expressed in terms of m1, a, F, mK and g

First, do a complete FBD and Newton’s Laws analysis of each mass, as follows:

(cable)

elevator

F

qm2 m1

x-analysis

SFx = m1ax

Fx – Fk.1 – FT.21 = m1ax

Fcosq – mkFN.1 – FT.21 = 0

y-analysis

SFy = m1ay

Fy + FN.1 – FG.1 = m1ay

Fsinq + FN.1 – m1g = m1ay

FBD of m1

FBD of m2

x-analysis

SFx = m2ax

FT.12 – Fk.2 = m2ax

FT.12 – mkFN.2 = 0

y-analysis

SFy = m2ay

FN.2 – FG.2 = m2ay

FN.2 – m2g = m2ay

coord. axes

y

x

FG.1

FN.1

F

qFT.21

Fk.1

coord. axes

y

x

FG.2

FN.2

FT.12

Fk.2

24

Now solve for m2, as follows:

I. Solve the y-force equation (Fsinq + FN.1 – m1g = m1ay) for FN.1: FN.1 = m1ay + m1g – Fsinq

II. Solve the x-force equation (Fcosq – mkFN.1 – FT.21 = 0) for FT.21: FT.21 = Fcosq – mkFN.1 = Fcosq – mk(m1ay + m1g – Fsinq) III. Solve for FT.12: By Newton’s Third Law, the magnitudes of FT.21 and FT.12 must be equal. FT.12 = FT.21 = Fcosq – mk(m1ay + m1g – Fsinq) IV. Solve the x-force equation (FT.12 – mkFN.2 = 0) for FN.2: FN.2 = FT.12/mk = Fcosq/mk – (m1ay + m1g – Fsinq)

= F(cosq/mk + sinq) – m1(ay + g)

V. Solve the y-force equation (FN.2 – m2g = m2ay) for m2: m2 = FN.2/(g + ay) = F(cosq/mk + sinq)/(ay + g) – m1

25

11. b. Walking through your neighborhood, you see Thor set down his hammer while working on the roof. The hammer slides, starting from rest, a distance d down the roof, which is angled q from the horizontal. The hammer leaves the roof from a height h, and lands a horizontal distance x from the point where it becomesaprojectile.Whatisthecoefficientofkineticfrictionbetweenhammerandtheroof?

You may consider these values as known: d, q, h, x, g. This is an ODAVEST problem, but keep in mind that you’re not being asked to actually solve for the final expression. In fact, you’re not being asked to do any math at all—not even any algebra. Rather, for the Solve step, you are to write a series of succinct instructions on how to solve this problem. Pretend that your instructions will be given to someone who knows math but not physics. And for the Test step, you should consider the situation and predict how the solution would change if the data were different—changing one data value at a time.

Objective: An object starts from rest and slides a distance d down a roof surface that is inclined at an angle q below the horizontal. The object leaves the roof at a height h above the ground and first impacts the ground at a horizontal distance x from the point where it left the roof. We need to determine mK (the coefficient of kinetic friction) between the object and the roof surface.

Data: vi = 0 The object began its slide at rest.

d The distance over which the object slid on the roof surface.

q The angle of incline of the roof surface, with respect to the horizontal.

h The vertical distance from the impact point to where the object left the roof.

x The horizontal distance from the impact point to where the object left the roof.

g The local free-fall acceleration magnitude.

Assumptions: Objects We will treat the hammer (object) as a point mass—having no extent or rotation.

Surfaces We will assume that the surfaces of both the roof and the hammer are uniform. We will also assume that the roof is planar.

Slide We assume that the slide was in a straight line and directly down the slope (along the steepest possible path).

Projectile We assume the g is constant for all relevant heights here. We will disregard any effects of wind or air drag.

Visual Reps.:

Free-body diagram of object sliding against kinetic friction down the roof slope:

Kinematics diagram (and “visual” data inventory) of the slide: FG.P1

FN.S1

FK.S1

q

Dx = d

vi = 0

vf

q

(x) (y)Dx = dvi.x = 0vf.x =ax = ?? Dt =

26

Kinematics diagram (and “visual” data inventory) of the projectile motion:

Equations: I. m1gsinq – mkFN.S1 = m1ax

II. FN.S1 – m1gcosq = 0

III. vf.x2 = 02 + 2axd

IV. x = vicosq(Dt)

V. –h = visinq(Dt) – (1/2)g(Dt)2

Solving: Solve IV for Dt. Substitute that result into V.

Solve V for Dt. Substitute that result into IV.

Solve IV for vi. Substitute that result as vf.x into III.

Solve III for ax. Substitute that result into I.

Solve II for FN.S1. Substitute that result into I.

Solve I for mk.

Testing: Dimensions: The solution would need to be unitless (just a numerical value).

Dependencies: In general, this number should be somewhere between 0 and 1 (or possibly a little above 1 if the roof is very sticky or rough).

A greater vi that produces the same projectile motion would imply that the roof slows down the slide a little more—a greater mk.

A greater slide distance d that produces the same projectile motion would imply that the acceleration is less (achieves same launch speed over a longer slide): a greater mk.

A higher fall, h, with the same x as before, would imply a slower launch speed: a greater mk.

Without explicitly solving the above equation set as prescribed, the effects of a greater roof angle q are not clear.

A greater x travel for the same fall height would imply a faster launch speed: a smaller mk.

Without explicitly solving the above equation set as prescribed, the effects of a greater value of g are not clear. It would affect the friction, the slide and the fall. (It’s possible, therefore, that it might even have no implications at all for mk.

(x) (y)Dx = x Dy = –hvi.x = vicosq vi.y = visinqvf.x = vf.y =ax = 0 ax = –g Dt =

viqi

vf

qfDx = x

Dy

= –

h

27

FBD of m

coord. axes

12. A 90-kg man boards an elevator and stands on an accurate bathroom scale (which just happens to be there). During his elevator ride, the scale reads a steady 700 N. Also during this ride, he happens to drop a coin from a height of 1 m above the elevator floor.Fromthemomenthereleasesthecoin,how longdoesittaketohittheflooroftheelevator?

x-analysis

SFx = max

0 = 0

y-analysis

SFy = may

FN – FG = may

FN – mg = may

The x-direction has no forces and thus zero acceleration. In the y-direction, ay is easily solved for:

ay = (FN – mg)/m = [700 – (90)(9.80)]/90 = –2.022 m/s2

This is the acceleration of the elevator floor, the man, and the coin in his hand—they all move as one—until the moment when he releases the coin. After that moment, the coin’s acceleration is –g. But the elevator floor’s acceleration continues to be ay (–2.022). Essentially then, the question becomes this: How long does it take the coin to “catch up” to the elevator floor as the both accelerate downward—at different rates?

Object A (coin): DyA = vA.i(Dt) + (1/2)(–g)(Dt)2

Object B (elevator floor): DyB = vB.i(Dt) + (1/2)(ay)(Dt)2

But DyA = DyB – 1 (It’s –1 because Dy downward is negative; A travels downward 1 meter more than B.)

And vA.i = vB.i (When the coin is released, its velocity is the same as the elevator floor’s.)

Substituting: vA.i(Dt) + (1/2)(–g)(Dt)2 = vA.i(Dt) + (1/2)(ay)(Dt)2 – 1

Solve for Dt: –(1/2)(ay)(Dt)2 – (1/2)(g)(Dt)2 = – 1

(Dt)2[(1/2)(ay) + (1/2)(g)] = 1

(Dt) = {1/[(1/2)(ay) + (1/2)(g)]}1/2

= {1/[(1/2)(–2.022) + (1/2)(9.80)]}1/2 = 0.507 s

(scale)

elevator

m (person)y

x

FN

FG