v T R x m Version PREVIEW Practice 7 carroll (11108) 1

(1)

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Atwood Machine 05

001 10.0 points

A 6.3 kg mass is attached to a light cord that

passes over a massless, frictionless pulley. The

other end of the cord is attached to a 3.4 kg

mass.

The acceleration of gravity is 9.8 m/s

2

.

5.6 m

ω

6.3 kg

3.4 kg

Use conservation of energy to determine the

final speed of the first mass after it has fallen

(starting from rest) 5.6 m .

Correct answer: 5.72842 m/s.

Explanation:

Let : m

₁

= 6.3 kg ,

m

₂

= 3.4 kg ,

and

ℓ = 5.6 m .

Consider the free body diagrams

6.3 kg

3.4 kg

T

m

1

g

m

2

g

a

Let the figure represent the initial

config-uration of the pulley system (before m

₁

falls

down).

From conservation of energy

K

i

+ U

i

= K

f

+ U

f

0 + m

₁

g ℓ = m

₂

g ℓ +

1

2 m

1

v

2

₊

1

2 m

2

v

2

(m

₁

−

m

₂

) g ℓ =

1

2 (m

1

+ m

2

) v

2

Therefore

v =

s

(m

₁

−

m

₂

)

(m

₁

+ m

₂

)

2 g ℓ

=

6.3 kg − 3.4 kg

6.3 kg + 3.4 kg

×

2 (9.8 m/s

2

)(5.6 m)

1/2

= 5.72842 m/s .

Frictionless Vertical Circle

002 (part 1 of 3) 10.0 points

A block of mass 0.5 kg is pushed against a

hor-izontal spring of negligible mass, compressing

the spring a distance of ∆x as shown in the

fig-ure. The spring constant is 276 N/m. When

released, the block travels along a frictionless,

horizontal surface to point B, the bottom of

a vertical circular track of radius 0.9 m, and

continues to move up the track. The speed

of the block at the bottom of the track is

15 m/s, and the block experiences an

aver-age frictional force of 5 N while sliding up the

track.

The acceleration of gravity is 9.8 m/s

2

.

m

k

R

v

_B

v

_T

T

B

∆

x

What is ∆x?

Correct answer: 0.638442 m.

(2)

Explanation:

From conservation of energy, the initial

po-tential energy of the spring is equal to the

kinetic energy of the block at B. Therefore,

we write

1

2 k (∆x)

2

₌

1

2 m v

2 B

∆x =

s

m v

2 B

k

=

s

(0.5 kg) (15 m/s)

2

(276 N/m)

= 0.638442 m .

003 (part 2 of 3) 10.0 points

What is the speed of the block at the top of

the track?

Correct answer: 11.54 m/s.

Explanation:

The change in the total energy of the block

as it moves from B to T is equal to the work

done by the frictional force

∆E = W

f

E

T

−

E

B

= W

f

.

The total energy at B is

E

B

=

1

2 m v

2 B

=

1

2 (0.5 kg) (15 m/s)

2

= 56.25 J .

The work done by the frictional force is

W

f

= −f π R

= −(5 N) (π) (0.9 m)

= −14.1372 J .

Therefore, the total energy at T is

E

T

= E

B

+ W

f

= 56.25 J + (−14.1372 J)

= 42.1128 J .

We can find now the speed of the block at T

from

1

2 m v

T 2

_{= E}

T

−

m g h

T

.

Since

v

T2

=

2 E

T

m

−

2 g h

T

,

=

2 (42.1128 J)

0.5 kg

−

2(9.8 m/s

2

) (1.8 m)

= 133.171 m

2

/s

2

,

then

v

T

=

q

133.171 m

2

_/s

2

= 11.54 m/s .

004 (part 3 of 3) 10.0 points

What is the centripetal acceleration of the

block at the top of the track?

Correct answer: 147.968 m/s

2

.

Explanation:

The centripetal acceleration at T is

a

c

=

v

2 T

R

=

(11.54 m/s)

2

0.9 m

= 147.968 m/s

2

.

Bead on a Loop the Loop 01

005 10.0 points

A bead slides without friction around a

loop-the-loop. The bead is released from a height

of 26.3975 m from the bottom of the

loop-the-loop which has a radius 9.999 m.

The acceleration of gravity is 9.8 m/s

2

.

26.3975 m

9.999 m

A

(3)

What is its speed at point A ?

Correct answer: 11.1996 m/s.

Explanation:

Let :

R = 9.999 m and

h = 26.3975 m .

From conservation of energy, we have

K

i

+ U

i

= K

f

+ U

f

0 + m g h =

m v

2

2 + m g (2 R)

v

2

= 2 g (h − 2 R) .

Therefore

v =

p2 g (h − 2 R)

=

r

2 (9.8 m/s

2

₎

h

_{26.3975 m − 2 (9.999 m)}

i

= 11.1996 m/s .

Block Jump Ramp 01

006 (part 1 of 3) 10.0 points

A block starts at rest and slides down a

fric-tionless track. It leaves the track horizontally,

flies through the air, and subsequently strikes

the ground.

b b b b_b b b b b b b b

441 g

3 .6

m

1 .8

m

x

v

What is the speed of the ball when it leaves

the track?

The acceleration of gravity is

9.81 m/s

2

.

Correct answer: 5.94273 m/s.

Explanation:

Let :

g = −9.81 m/s

2

,

m = 441 g ,

and

h

₁

= 1.8 m .

b b b b_b b b b b b b b

m

h

1

h

2

3.6 m

v

Choose the point where the block leaves the

track as the origin of the coordinate system.

While on the ramp,

K

b

= U

t

1

2 m v

2 x

= −m g h

1

v

2_x

= −2 g h

₁

v

x

=

p

−

2 g h

₁

=

q

−

2 (−9.81 m/s

2

_{) (1.8 m)}

= 5.94273 m/s .

007 (part 2 of 3) 10.0 points

What horizontal distance does the block

travel in the air?

Correct answer: 3.6 m.

Explanation:

Let :

h

₂

= −1.8 m .

With the point of launch as the origin,

h

₂

=

1

2 g t

2

t =

s

2 h

₂

g

.

Thus

x = v

x

t = v

x

s

2 h

₂

g

= (5.94273 m/s)

s

2 (−1.8 m)

−

9.81 m/s

2

= 3.6 m .

(4)

008 (part 3 of 3) 10.0 points

What is the speed of the block when it hits

the ground?

Correct answer: 8.40428 m/s.

Explanation:

Let :

h = 1.8 m .

Now choose ground level as the origin.

nergy conservation gives us

K

f

= U

i

1

2 m v

2 f

= −m g h

(7)

v

f

=

p

−

2 g h

=

q

−

2 (−9.81 m/s

2

_{) (1.8 m)}

= 8.40428 m/s .

Alternate Solution:

v

y

=

p

−

2 g h

₂

(9)

=

q

−

2 (−9.81 m/s

2

_{) (1.8 m)}

= 5.94273 m/s ,

so

v

f

=

q

v

_x2

_{+ v}

_y2

₍₁₀₎

=

q

(5.94273 m/s)

2

_{+ (5.94273 m/s)}

2

= 8.40428 m/s .

Serway CP 05 63

009 (part 1 of 3) 10.0 points

Two objects are connected by a light string

passing over a light frictionless pulley as

shown in the figure. The 4.51 kg object is

released from rest at a point 4.51 m above the

floor.

The acceleration of gravity is 9.8 m/s

2

.

4.51 m

ω

4.51 kg

2.17 kg

Find the speed of each object just as they

pass each other.

Correct answer: 3.93478 m/s.

Explanation:

Let : m

₁

= 4.51 kg ,

m

₂

= 2.17 kg ,

and

h = 4.51 m .

Consider the free body diagrams

2.17 kg

4.51 kg

T

m

2

g

m

1

g

a

When they meet both are

1

2 h from the

floor, and they have the same speed since

they are connected by the string. Applying

conservation of energy,

(K + U

g

)

_i

= (K + U

g

)

_f

m

₁

g h =

1

2 m

1

v

2

₊

1

2 m

2

v

2

+ m

₁

g

1

2 h

+ m

₂

g

1

2 h

2 m

₁

g h = (m

₁

+ m

₂

) v

2

(5)

+ (m

₁

+ m

₂

) g h

m

₁

g h − m

₂

g h = (m

₁

+ m

₂

) v

2

v

2

=

(m

1

−

m

2

) g h

m

₁

+ m

₂

.

Thus

v =

s

(m

₁

−

m

₂

) g h

m

₁

+ m

₂

=

s

(4.51 kg − 2.17 kg) (9.8 m/s

2

_{) (4.51 m)}

2.17 kg + 4.51 kg

= 3.93478 m/s

010 (part 2 of 3) 10.0 points

b) What is the speed of the each object at the

instant before the 4.51 kg hits the floor?

Correct answer: 5.56463 m/s.

Explanation:

Given : y

_2f

= y

_1f

= h

y

_1f

= 0

Conservation of mechanical energy gives

m

₁

g h =

1

2 (m

1

+ m

2

) v

2 f

+ m

2

g h

v

2_f

=

2 (m

1

−

m

2

) gh

m

₁

+ m

₂

v

f

=

s

2 (m

₁

−

m

₂

) g h

m

₁

+ m

₂

=

s

2 (4.51 kg − 2.17 kg)

2.17 kg + 4.51 kg

×

q

(9.8 m/s

2

_{) (4.51 m)}

= 5.56463 m/s .

011 (part 3 of 3) 10.0 points

c) How much higher does the 2.17 kg object

travel after the 4.51 kg object hits the floor?

Correct answer: 1.57985 m.

Explanation:

The 2.17 kg mass becomes a

projec-tile launched straight upward with v

iy

=

5.56463 m/s.

v

_{f y}2

= v

2_iy

−

2 g ∆y = 0

at the maximum height, so

∆y

max

=

v

2 iy

2 g

=

(5.56463 m/s)

2

2 (9.8 m/s

2

₎

= 1.57985 m .

Sliding Down a Dome

012 (part 1 of 3) 10.0 points

A small box of mass m is at the top of a

spherical dome with radius R. Starting from

rest after a slight push, the box slides down

along the frictionless spherical surface (see

figure). Ignore the initial kinetic energy

(ob-tained from the slight push).

θ

R

m

v

Apply the principle of conservation of

en-ergy to select the correct equation that relates

the speed v of the box to the polar angle θ and

the radius R while the box is sliding down on

the surface.

1. v =

p2 g R (1 − cos θ) correct

2. v =

pg R

3. v =

p2 g R sin θ

4. v =

p2 g R

5. v =

pg R cot θ

6. v =

p2 g R (1 − sin θ)

7. v =

pg R sin θ

(6)

8. v =

r

g R (1 + tan θ)

2 9. v =

p2 g R cos θ

10. v =

pg R tan θ

Explanation:

Basic Concepts: Potential Energy,

Ki-netic Energy, Centripetal Force

θ

R

m

v

θ

mg

Since the surface is frictionless, the

me-chanical energy of the box is conserved, so the

initial mechanical energy is the same as the

final mechanical energy

m g R +

1

2 m v

2 i

= m g R cos θ +

1

2 m v

2

_.

It starts from rest (the push is “slight”) so

v

i

= 0, and we find

m g R (1 − cos θ) =

1

2 m v

2

v =

p2 g R (1 − cos θ) .

013 (part 2 of 3) 10.0 points

Select the inequality relation which

corre-sponds to the condition that the small box

will stay on the surface.

1. v

2

R < g (1 + cos θ)

2. v

2

R

> g sin θ

3. v

2

R

< g cos θ correct

4. 2 v

2

_{> g R (1 + cot θ)}

5. v

2

2 < g R (1 − sin θ)

6. v

2

R

< 2 g tan θ

7. v > g R cot θ

8. v

2

R

< g sin θ

9. v < g R sin θ

10. v

2

sin θ

2 > g (1 + cos θ)

Explanation:

The radial acceleration of the box is v

_R2

while it is on the surface. So the net radial

force on the box is

m v

2

R

. The normal force the

surface exerts on the box can only be outward

so gravity must supply the total force inward,

with any extra counteracted by the normal

force, which must be positive. Thus

m v

2

R

= m g cos θ − N ≤ m g cos θ .

Therefore,

v

2

R

≤

g cos θ .

014 (part 3 of 3) 10.0 points

Find the critical angle at which the box leaves

the surface of the dome.

1. θ = 36.9

◦

2. θ = 41.7

◦

3. θ = 48.2

◦

_correct

4. θ = 49.7

◦

5. θ = 46.5

◦

6. θ = 39.3

◦

7. θ = 51.3

◦

8. θ = 34.1

◦

9. θ = 44.1

◦

10. θ = 38.6

◦

Explanation:

(7)

The box leaves the surface when the normal

force is zero. At this instant the component

of the gravity force along the normal to the

surface will be equal to the mass times the

centripetal acceleration

v

2

R

.

From the

dia-gram, we see that

m v

2

R

= m g cos θ .

(1)

From Part 1:

1

2 m v

2

_{= m g R (1 − cos θ)}

m v

2

R

= 2 m g (1 − cos θ) .

(2)

Putting Eqs. (1) and (2) together, we have

m g cos θ = 2 m g (1 − cos θ)

3 m g cos θ = 2 m g

cos θ =

2

3 θ = 48.2

◦