12 Solution Sheets Ch 5

62

Chapter 5 · Motion Near the Earth's Surface Section 5.1 Practice 1.

a)

(page 155)

Fg = mg

Fe -

980 N = (100 kg)(0.30 m/s 2 )

Fe -

980 N = 30 N

= (50 kg}(9.8 N/kg) = 4.9 b)

Fg

X

lO 2 N

= mg

Fe = 1 010 N

= (0.100 kg)(9.8 N/kg)

Fe

= 0.98 N 2.

g = 1.6 N/kg a)

= 1 010 N [up]

c)

Fg = mg

= (50 kg}(1.6 N /kg)

b)

Fg

= 80 N = mg = (0.100 kg}(1.6 N/kg) = 0.16 N

Practice (page 158) 1.

a}

-?

at rest, a = 0

and

-?

Fnel = 0

Fg = mg

= (100 kg)(9.8 N/kg) Fe- 980 N = -15 N

= 980 N -?

-?

-?

Fe = 965 N

Fnet = Fe + Fg

0 = Fe -

980 N

-?

Fe = 965 N [up]

Fe= +980 N -?

Fe = 980 N [up]

d)

b)

If v is uniform,

Fe

-?

a = 0, Fnet = 0, o o o

Fe = 980 N [up] as in (a).

63 e)

If falling freely, there is no force of

2.

c)

the elevator on the passenger, as seen below. -'>

ma F -

e

980 N Fe

= -(100 =0N

T

kg)(9.8 m/s 2 )

2. T

a•O -'>

~et

-'>

=T +

-+

~

Fg

T - 19 600 N

T

= 2.0

X

= ma =0

10 4 N

d) Fg

mg

(2000 kg)(9.8 N/kg) 19 600 N a)

-'>

-'>

-'>

=

T -

19 600 N

T

+

-'>

18

Fnet

ma

=

0

=0

T = 19 600 N T = 2.0

X

10 4 N ~

Fnet

b)

=

~

T

+

~

~

Fg

= ma

= ma T- 19 600 N =(2000 kg)(l.O m/s

T- mg T

a•O

T- 19 600 N

=2000 N

T

= 21600 = 2.2

e)

-'>

ma T- 19 600 N = 0 T = 19 600 N

= 2.0

X

10 4 N

0

X

N

10 4 N

2 )

64 2.

e)

(cont.)

_, _, _, Fnet = T + Fg T- rng = rna T -

19 600 N

T -

19 600 N

=

_, rna

aI

=

Fnet

Fnet

__ I

rn I

a)

Fg

=

T

(2000 kg)(- 1.0 m/s 2

=-

)

=

2000 N

T

17 600 N

T

1.8

X

g_

rn 2 T

=

5.0 kg

a

10 4 N

5.0 kg

rng

!_g_

rn=

g 4

-~

98 N 9.8 N/kg _, 4

3.0 T

10 kg

T

~ + Fg =rna

Fnet

a

=

29.4 N-T 3.0 kg 147 N- 5.0 T

= 18 N ~

----·r0gl

(10 kg)a

= -0.50

a2

40kg

93 N - 98 N = rna

=

29.4 N-T 3.0 kg

2.

Vertical components:

-5.0 N

rn 2 F- T

rn I

T

3.

__ 2

a2

N/kg

= -0.50 m/s 2 b)

Acceleration is down, since value of a is negative.

Practice (page 161) 1.

5.0kg

~

----·~

Fg

rnz g (15 kg)(9.8 N/kg) 147 N

aI

!

Fnet rn I

Fnet

__ I

a2

40 kg

aI T

40 kg

Fg

M2 g

a2 147 N-T 15 kg

15 T = 5880 N- 40 T

(3.0 kg)(9 .8 N /kg) T = 107 N

29.4 N

rn 2 147 N-T 15 kg

T a2

__ 2

65

=

a

=

T

Section 5.4

mI Practice (page 165)

107 N 40 kg

= 2.67

m/s 2

1.

p

-

F

:.1__

k - FN

= 2.7 m/s 2 3.

= ---=-5.::...0-=.N_c___ _ (20 kg)(9.8 N/kg)

==

3 U KQ

5.0 kg

2.

0.26

Ff = f.ik FN

= ~"k mg =

0.30(15 kg)(9.8 N/kg)

= 44 N 3.

Fgl =m 18 = (3.0 kg)(9.8 N/kg)

=

F

29.4 N

Fg2 =m 2 g = (5.0 kg)(9.8 N/kg)

= aI

= =

49 N

Fnet

mI T- 29.4 N

Ff

Fnet

a2

pkFN

= 0.12(2.0 kg)(9.8 N/kg)

m2

=

49 N-T 5.0 kg

3.0 kg

=

2.35 N

Fnet = ma

a I = a2

= ma = (2.0 kg)a = 0.83 m/s

F- Ff 4.0 N - 2.35 N

T- 29.4 N 49 N-T = 3.0 kg 5.0 kg 5.0 T- 147 N = 147 N- 3.0 T

a 4.

8 T = 294 N T = 37 N

aI

=

T - 29.4 N

3.0 kg

37 _ NN = .::c__ _29.4 ...c_:_-'----'--' 3.0 kg

= 2.5 m/s 2 -

-

----"

2

66 Fx

=

b)

F cos8

(1 00 N)( cos 55 •)

-

Fx

=

57 N

FY

=

F sin 55.

82 N Since v is constant, Fncl

=

ma

=

0 and

=

FN

Fg cos8

Ff = FX = 57 N

(98 N)(cos10•)

But

J1 k

=

96.5 N

!1_

Jl.k FN

FN

(0.10)(96.5 N)

57 N

9.65 N

82 N 0.70

Taking down the slope as positive

5. F = 17 N -

9.65 N

= 7.35 N a -

F m

_ 7.35 N - 10 kg

= 0.735 m/s bod=

V

1

bo(

- 0 20ma)

Fg = mg

b.t

= (10 kg)(9.8 N/kg)

98 N F

Fg sin8

= (98 N)(sin 10 •) = 17 N F m

a=-

17 N 10 kg

= 1.7m/s 2

6.

+

+~

2

a(b.t) 2

1 2(0.735 m/s 2 )(b.t) 2

= 7.38 s

or 7.4 s

67 FN

= Fy = mgcos20°

mg

Fg I

(40 kg)(9.8 N/kg)

Fx = mg sin 20 o a)

392 N

Ff = f-lk FN

but

= O.lO(mg cos 20 °)

FN =Fg,

)

)

Fnel = Fx

=

Fnet,x

+

= 392 N

FN

= 0.093 mg N -)

mg sin 20 o

Pk FN

Ff

Ff

(0.35)(392 N)

0.093 mg

-

137N

= 0.34 mg- 0.093 mg

[1]

= (0.,34- 0.093)(9.8 N/kg)m N

T - 137 N

F nel,,

rna

= 2.4 m N

I

(40 kg)a

Fnel,2

(2] 2.4 m

F g2

b)

v

T = m 2 a2 T

147 N -

m 2.4 m/s

I

= (15 kg)a

2

2

Combining [1] and (2] and given that

= v I + a D.l

= 0 + (2.4 m/s = 19.4 m/s = 19 m/s

2

= a2 = a

a,

)(8.0 s)

(one rope)

T- 137 N = 40a T = 15a

147 N -

lON

Note that the acceleration and the final speed of the skier do not depend on the mass of the skier.

= 55a =

a

0.18 m/s

2

Section 5.6

7. 40kg

~

Practice (page 169)

----~

1.

a)

Fg

=

mg

(60 kg)(9.5 N /kg) 5.7 b)

X

10< N

ag = g = 9.5 m/s 2 v2

=

vI 0

+ a!J.t

+

(9.5 m/s 2 )(10 s)

T

= 95 m/s

2.

Fs

g =m 1830 N 70.0 kg 26.1 N/kg