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jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The Joule and the kilowatt-hour are both units of energy. 19.2 kW · h is equivalent to how many Joules?

1

Correct answer: 455.786 W. Explanation: Let : F = 709 N , d = 9 m , and t = 14 s . The power expended is

7

Correct answer: 6.912 × 10 J. Explanation: 1 W = 1 J/s

Fd W = t t (709 N) (9 m) = 14 s

P =

1000 W 3600 s · kW 1h = 6.912 × 107 W · s

19.2 kW · h = (19.2 kW · h) ·

= 6.912 × 107 J . 002 10.0 points A cheerleader lifts his 59.9 kg partner straight up off the ground a distance of 0.557 m before releasing her. The acceleration of gravity is 9.8 m/s2 . If he does this 27 times, how much work has he done? Correct answer: 8828.19 J. Explanation: The work done in lifting the cheerleader once is W1 = m g h = (59.9 kg)(9.8 m/s2 )(0.557 m) = 326.97 J .

= 455.786 W . 004 10.0 points You leave your 75 W portable color TV on for 7 hours each day and you do not pay attention to the cost of electricity. If the dorm (or your parents) charged you for your electricity use and the cost was $0.1 /kW · h, what would be your monthly (30 day) bill? Correct answer: 1.575 dollars. Explanation: Let : P = 75 W and t = 7 h/day , The energy consumed in each day is W =Pt

The work required to lift her n = 27 times is

= (75 W) (7 h/day) ·

W = n W1 = (27)(326.97 J) = 8828.19 J.

= 0.525 kW · h/day.

kW 1000 W

In 30 days, you would use 003 10.0 points A student weighing 709 N climbs at constant speed to the top of an 9 m vertical rope in 14 s. What is the average power expended by the student to overcome gravity?

(30 day) (0.525 kW · h/day) = 15.75 kW · h, which would cost you (15.75 kW · h) ($0.1/kW · h) = $1.575 .

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) 005 10.0 points A 105 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 87 m below the rim. A student (majoring in linguistics and physics) decides to perform a rescue/experiment using a nearby horse. After lowering a rope to her fallen hero and attaching the other end to the horse, the student measures how long it takes for the horse to pull the fallen physicist to the rim of the Grand Canyon. The acceleration of gravity is 9.8 m/s2 . If the horse’s output power is truly 1 horsepower (746 W), and no energy is lost to friction, how long should the process take? Correct answer: 120.004 s. Explanation: The work against gravity which is necessary to raise the professor to the rim of the canyon is given by W = m g ∆h = (105 kg) (9.8 m/s2 ) (87 m) = 89523 J . If the horse’s power output is one horsepower, then the the time for the horse to provide the necessary work is given by tup =

W

Poutput 89523 J = 746 W = 120.004 s .

006 10.0 points Potential energy and kinetic energy are forms of what kind of energy?

2

5. heat Explanation: 007 10.0 points A Joule is a measure of 1. density. 2. distance. 3. volume. 4. momentum. 5. energy. correct Explanation: 008 10.0 points Shawn and his bike have a total mass of 58.7 kg. Shawn rides his bike 0.77 km in 13.9 min at a constant velocity. The acceleration of gravity is 9.8 m/s2 . What is Shawn’s kinetic energy? Correct answer: 25.0184 J. Explanation: Shawn’s velocity is v=

d . t

Thus his kinetic energy is 1 K = mv2 2  0.77 km 1000 m 1 min 2 1 · · = (58.7 kg) 2 13.9 min 1 km 60 s = 25.0184 J .

3. mechanical correct

009 10.0 points Tim, with mass 71.6 kg, climbs a gymnasium rope a distance of 2 m. The acceleration of gravity is 9.8 m/s2 . How much potential energy does Tim gain?

4. electromagnetic

Correct answer: 1403.36 J.

1. chemical 2. nuclear

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) Explanation: Let : m = 71.6 kg , h = 2 m , and 2

g = 9.8 m/s . Potential energy is U = mg h = (71.6 kg) (9.8 m/s2 ) (2 m) = 1403.36 J . 010 (part 1 of 3) 10.0 points A 40.0 kg child is in a swing that is attached to ropes 2.30 m long. The acceleration of gravity is 9.81 m/s2 . Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions: a) when the ropes are horizontal.

Explanation: Given: θ = 36.0◦ Solution: The child is a distance ℓ cos θ from the attachment point, so h2 = ℓ − ℓ cos θ and Ug = mgh2 = mgℓ(1 − cos θ ) = (40 kg)(9.81 m/s2 )(2.3 m) · (1 − cos 36◦ ) = 172.366 J 012 (part 3 of 3) 10.0 points c) at the bottom of the circular arc. Correct answer: 0 J. Explanation: Solution: The child is at the lowest point, so h3 = 0, and Ug = mgh3 = (40 kg)(9.81 m/s2)(0 m) =0J

Correct answer: 902.52 J. Explanation: Basic Concept:

013

Ug = mgh Given: ℓ = 2.30 m m = 40.0 kg g = 9.81 m/s2 Solution: The child is at a height h1 = ℓ from the lowest point, so Ug = mgh1 = (40 kg)(9.81 m/s2)(2.3 m) = 902.52 J 011 (part 2 of 3) 10.0 points b) when the ropes make a 36.0◦ angle with the vertical. Correct answer: 172.366 J.

3

10.0 points

A block of mass m slides on a horizontal frictionless table with an initial speed v0 . It then compresses a spring of force constant k and is brought to rest. v k m m µ=0 How much is the spring compressed x from its natural length? mk g r m 2. x = v0 correct k v2 3. x = 0 2g k 4. x = v0 gm

1. x = v0

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) r

Anyone who checks to see if the units are correct should get this problem correct. 014 10.0 points A bead slides without friction around a loopthe-loop. The bead is released from a height of 11.6 m from the bottom of the loop-theloop which has a radius 3 m. The acceleration of gravity is 9.8 m/s2 .

From conservation of energy, we have Ki + Ui = Kf + Uf m v2 0 + mg h = + m g (2 R) 2 v 2 = 2 g (h − 2 R) . Therefore v= =

p

2 g (h − 2 R) r h

2 (9.8 m/s2 ) 11.6 m − 2 (3 m)

= 10.4766 m/s .

015 (part 1 of 3) 10.0 points A block starts at rest and slides down a frictionless track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground. 570 g v

x

A 3m 11.6 m

What is the speed of the ball when it leaves the track? The acceleration of gravity is 9.81 m/s2 . Correct answer: 7.00357 m/s. Explanation:

What is its speed at point A ? Correct answer: 10.4766 m/s. Explanation:

i

2.4 m

1 1 m v02 = Ei = Ef = k x2 , or 2 2 m 2 2 , therefore x = v k r0 m . x = v0 k

Let : R = 3 m and h = 11.6 m .

4.9 m

k m mg 6. x = v0 k m 7. x = v0 kg s k 8. x = v0 mg r mg 9. x = v0 k 2 v 10. x = 0 2m Explanation: Total energy is conserved (no friction). The spring is compressed by a distance x from its natural length, so 5. x = v0

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Let : g = −9.81 m/s2 , m = 570 g , and h1 = 2.5 m .

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3)

h1

m

5

017 (part 3 of 3) 10.0 points What is the speed of the block when it hits the ground?

h

v

h2

Correct answer: 9.805 m/s. Explanation: x

Choose the point where the block leaves the track as the origin of the coordinate system. While on the ramp, Kb = Ut 1 m vx2 = −m g h1 2 vx2 = −2 g h1 p vx = −2 g h1 q = −2 (−9.81 m/s2 ) (2.5 m)

Let :

h = 4.9 m .

Now choose ground level as the origin. Energy conservation gives us

Kf = Ui 1 m v2f = −m g h 2 p vf = −2 g h q = −2 (−9.81 m/s2 ) (4.9 m) = 9.805 m/s .

= 7.00357 m/s .

Alternate Solution: 016 (part 2 of 3) 10.0 points What horizontal distance does the block travel in the air?

vy = =

Correct answer: 4.89898 m. Explanation: h2 = −2.4 m .

Let :

With the point of launch as the origin, 1 h2 = g t2 2 s

2 h2 . g

t=

Thus x = vx t = vx

s

2 h2 g s

= (7.00357 m/s) = 4.89898 m .

2 (−2.4 m) −9.81 m/s2

p

q

−2 g h2

−2 (−9.81 m/s2 ) (2.4 m)

= 6.86207 m/s , so q vf = vx2 + v 2y q = (7.00357 m/s)2 + (6.86207 m/s)2 = 9.805 m/s .

018 (part 1 of 3) 10.0 points A block is pushed against the spring with spring constant 9.1 kN/m (located on the lefthand side of the track) and compresses the spring a distance 4.5 cm from its equilibrium position (as shown in the figure below). The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.81 m/s2 .

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3)

6

522 g

4.5 cm

9.81 m/s2

9.1 kN/m

2.4 m

Solution: Using Eqs. 1, 2, and 3, we have v

1 1 m vx2 = k d2 2 2

(4)

k d2 m r k d2 vx = m s (9100 N/m) (0.045 m)2 = 0.522 kg

vx2 =

x What is the speed v of the block when it leaves the track? Correct answer: 5.94153 m/s. Explanation:

(5) (6)

= 5.94153 m/s .

Let : g = 9.81 m/s2 , m = 0.522 kg , k = 9100 N/m , vx = v , and

019 (part 2 of 3) 10.0 points What is the horizontal distance x the block travels in the air? Correct answer: 4.15608 m.

d = 0.045 m .

Explanation: At ∆y = h = −2.4 m (below the jump off height),

5.94 m/s m g

k

h

d

1 h = − g t2 s2 −2 h . t= g

4.16 m Basic Concepts: chanical Energy

Conservation of Me-

Ui = Uf + Kf ,

(1)

since vi = 0 m/s. K=

1 m v2 2

(2)

1 Us = k d2 . (3) 2 Choosing the point where the block leaves the track as the origin of the coordinate system, ∆x = vx ∆t 1 ∆y = − g ∆t2 , 2 since axi = 0 m/s2 and vyi = 0 m/s.

(7) (9)

Since x = vx t , therefore using Eq. 6 and 8, we have x = vx t s r k d2 −2 h = g m s 2 k d2 h = − mg  2 (9100 N/m) = − 0.522 kg

(0.045 m)2 (−2.4 m) × 9.81 m/s2 = 4.15608 m .

(10)

1/2

020 (part 3 of 3) 10.0 points

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) What is the total speed of the block when it hits the ground?

Correct answer: 20.0938 m/s.

Correct answer: 9.07688 m/s.

Explanation:

Explanation: Basic Concept: v=

Kf = Ui , since vi = 0 m/s and hf = 0 m. 1 Solution: Using m vy2 = m g h , we have 2 p vy = −2 g h (11) q = −2 (9.81 m/s2 ) (−2.4 m) = 6.86207 m/s , so q (12) vf = vx2 + vy2 q = (5.94153 m/s)2 + (6.86207 m/s)2 = 9.07688 m/s .

021 (part 1 of 2) 10.0 points While skiing in Jackson, Wyoming, your friend Ben (of mass 72.9 kg) started his descent down the bunny run, 20.6 m above the bottom of the run. If he started at rest and converted all of his gravitational potential energy into kinetic energy, what is Ben’s kinetic energy at the bottom of the bunny run? Correct answer: 14717.1 J.

s

2K m 2 (14717.1 J) 72.9 kg

= 20.0938 m/s .

023 (part 1 of 4) 10.0 points A 162 g particle is released from rest at point A along the diameter on the inside of a frictionless, hemispherical bowl of radius 47.5 cm. The acceleration of gravity is 9.8 m/s2 .

Calculate its gravitational potential energy at point A relative to point B . Correct answer: 0.75411 J. Explanation: The potential energy at A is

Explanation: Let : m = 72.9 kg , g = 9.8 m/s2 , h = 20.6 m .

=

r

and

Kbottom = Utop = mg h = (72.9 kg) (9.8 m/s2) (20.6 m) = 14717.1 J .

UA = m g hA = (162 g)(0.001 kg/g)(9.8 m/s2 ) × (47.5 cm)(0.01 m/cm) = 0.75411 J .

024 (part 2 of 4) 10.0 points Calculate its kinetic energy at point B . Correct answer: 0.75411 J.

022 (part 2 of 2) 10.0 points What is his final velocity?

7

Explanation:

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) The kinetic energy at B is equal to the potential energy at A, since all the energy went to kinetic energy. KB = UA = 0.75411 J . 025 (part 3 of 4) 10.0 points Calculate its kinetic energy at point C at 2R . height 3

Basic Concepts: Conservation of Mechanical Energy Ui = Kf since vi = 0 m/s and hf = 0 m. 1 K = mv2 2 Ug = mgh hi = ℓ − ℓ cos θ Given: ℓ = 31.0 m θ = 36.0◦ g = 9.81 m/s2

Correct answer: 0.25137 J. Explanation: The kinetic energy at C is KC = m g (hA − hC )   2R = mg R − 3 UA = 3 0.75411 J = 3 = 0.25137 J .

Solution: 1 mghi = mv2f 2 2 = 2gh vf i = 2gℓ(1 − cos θ ) = 2(9.81 m/s2 )(31 m)(1 − cos 36◦) = 116.16 m2 /s2 so that vf =

026 (part 4 of 4) 10.0 points Calculate its potential energy at point C .

q

116.16 m2 /s2

= 10.7777 m/s

Correct answer: 0.50274 J.

028 (part 2 of 2) 10.0 points b) pushes off with a speed of 6.00 m/s?

Explanation: The potential energy at C is

Correct answer: 12.3353 m/s.

UC = UA − KC = 0.75411 J − 0.25137 J = 0.50274 J . 027 (part 1 of 2) 10.0 points Tarzan swings on a 31.0 m long vine initially inclined at an angle of 36.0 ◦ with the vertical. The acceleration of gravity if 9.81 m/s2 . What is his speed at the bottom of the swing if he a) starts from rest? Correct answer: 10.7777 m/s. Explanation:

8

Explanation: Basic Concept: Ui + Ki = Kf since hf = 0 m. Given: vi = 6.00 m/s Solution: 1 1 2 mv f = mghi + mvi2 2 2 2 vf = 2gℓ(1 − cos θ) + vi2 = 2(9.81 m/s2 )(31 m) · (1 − cos 36◦ ) + (6 m/s)2 = 152.16 m2/s2

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) so that vf =

q

152.16

A block of mass m slids up a ramp making an angle θ with the horizontal. The block has 1 an initial KE of m v20 as it starts up the 2 ramp and travels a distance L along the ramp before coming to momentary rest.

m2/s2

= 12.3353 m/s 029

10.0 points

A 200 kg block is released at a 4.2 m height as shown. The track is frictionless. The block travels down the track, hits a spring of force constant k = 1623 N/m . The acceleration of gravity is 9.8 m/s2 . 200 kg 4.2 m

1623 N/m x

Determine the compression of the spring x from its equilibrium position before coming to rest momentarily. Correct answer: 3.18499 m. Explanation:

L

From conservation of energy ∆E = Wf

How much work did kinetic friction do on the block between its starting point and the point it came to momentary rest?   v 20 1. m g L sin θ − correct 2 2. Zero, since friction is perpendicular to mg 3. m g L tan θ 4. m g (sin θ − cos θ)

6. m g L sin θ 1 7. m v02, since the block comes to rest: 2 W = ∆KE 8. m g L cos θ Explanation:

k x2 −mg h = 0. 2

L θ

Hence, x= =

r

s

2mg h k 2 (200 kg) (9.8 m/s2) (4.2 m) (1623 N/m)

= 3.18499 m . 030

v0

5. m g L

Let : m = 200 kg , k = 1623 N/m , h = 4.2 m , and g = 9.8 m/s2 .

or

9

10.0 points

L sin θ

If we take the gravitational potential energy to be zero at the point where the block has its initial speed, then from E = K + Ug Ei =

1 m v02 + 0 and 2

Ef = 0 + m g L sin θ .

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) The work done by friction, a nonconservative force, is thus   v02 . Wf = Ef − Ei = m g L sin θ − 2 031 10.0 points A 16.9 kg block is dragged over a rough, horizontal surface by a constant force of 81.6 N acting at an angle of 29 ◦ above the horizontal. The block is displaced 95.4 m, and the coefficient of kinetic friction is 0.176. The acceleration of gravity is 9.8 m/s2 . 6N 81. ◦ 29

10

= F sx cos θ = (81.6 N) (95.4 m) cos 29◦ = 6808.6 J . To find the frictional force, Ff riction = µ N , we need to find N from vertical force balance. Note that N is in the same direction as the y component of F and opposite the force of gravity. Thus F sin θ + N = m g so that N = m g − F sin θ . Thus the friction force is

16.9 kg

~ Ffriction = −µ N ˆı = −µ (m g − F sin θ)ˆı .

µ = 0.176 The work done by friction is then If the block was originally at rest, determine its final speed. Correct answer: 23.5641 m/s. Explanation: Let : m = 16.9 kg , F = 81.6 N , sx = 95.4 m , µ = 0.176 , and g = 9.8 m/s2 .

Wµ = F~friction · ~s = −|fµ | |s| = −µ (m g − fµ sin θ) sx h = −(0.176) (16.9 kg) (9.8 m/s2 ) i ◦ −(81.6 N) sin 29 (95.4 m) = −2116.59 J .

The net work done on the block is equal to the change in kinetic energy, so 1 m v2 − 0 = WF + Wµ 2

Consider the force diagram N F

θ fk

r

2 [WF + Wµ ] v h m i u u 2 (6808.6 J) + (−2116.59 J) t = (16.9 kg)

v=

= 23.5641 m/s . mg ~ · ~s, where ~s is the distance Work is W = F traveled. In this problem ~s = 5ˆı is only in the x direction. ⇒ WF = Fx sx

032 10.0 points The spring has a constant of 10 N/m and the frictional surface is 0.4 m long with a coefficient of friction µ = 1.56 . The 2 kg block depresses the spring by 29 cm, then is

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) released. The first drop is 1 m and the second is 2 m. The acceleration of gravity is 9.8 m/s2 .

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the spring, then it gains kinetic energy by dropping a distance h1 , then it loses kinetic energy by doing work against friction. Us + Ug − Wfr = Kf =

1 m v2 2

Since 2 (Us + Ug − Wfr ) m 2 (0.4205 J + 19.6 J − 12.2304 J) = 2 kg 2 2 = 7.7901 m /s

v2 =

How far from the bottom of the cliff does it land?

then v=

Correct answer: 1.78315 m. Explanation: Basic concepts The potential energy of a spring is

Kinematics gives the time for the mass to drop a distance h2 1 2 gt 2 2 h2 t2 = g s 2 h2 t= g s 2 (2 m) = (9.8 m/s2) = 0.638877 s .

1 k x2 2 1 = (10 N/m) (29 cm)2 2 = 0.4205 J .

h2 =

Gravitational potential energy is Ug = m g h = (2 kg) (9.8 m/s2) (1 m) = 19.6 J .

Wfr = µ m g L = (1.56) (2 kg) (9.8 m/s2 ) (0.4 m) = 12.2304 J .

Thus the mass will land a distance s = vt = (2.79108 m/s) (0.638877 s) = 1.78315 m . from the bottom of the cliff.

Kinetic energy is K=

7.7901 m2 /s2

= 2.79108 m/s .

Us =

Work done against friction on a flat surface is

q

1 m v2 . 2

The height of the cliff determines how long it takes for the mass to reach the bottom. Energy considerations give us the horizontal velocity as the mass leaves the cliff top. The mass gets its initial kinetic energy from

033 10.0 points A solid metal ball and a hollow plastic ball of the same external radius are released from rest in a large vacuum chamber. When each has fallen 1 m, they both have the same 1. inertia.

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3)

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What average ...