Physics, 6th Edition
Chapter 7 Newton’s Second Law
Chapter 7. Newton’s Second Law Newton’s Second Law 7-1. A 4-k !ass !ass is acted acted on "y a res#$tant res#$tant %orce o% o% &a' 4 N, &"' ( N, and &c' 1) N. *hat are the the res#$tin acce$erations+ &a' a =
4N
= 1 !s)
4 k
&"' a =
(N 4 k
= ) !s)
&c' a
=
1)N 4 k
= !s)
7-). A constant constant %orce %orce o% ) N acts on a !ass o% &a' ) k, k, &"' 4 k, and &c' &c' 6 k. *hat are are the res#$tin acce$erations+ &a' a =
)N
= 1 !s)
) k
&"' a =
)N 4 k
=
&c' a
/ !s)
)N
=
6 k
= . !s)
7-. A constant %orce o% 6 $" acts on each o% three three o"0ects, prod#cin acce$erations acce$eration s o% 4, (, and 1) N. *hat are the !asses !asses++ m=
6 $"
= 1/ s$#s
4 %t s)
m=
6 $" ( %t s)
= 7. 7./ s$#s
m=
6 $" 1) %t s)
=
/ s$#s
7-4. *hat res#$tant %orce is necessary to ie a 4-k ha!!er an acce$eration o% 6 !s)+ F = ma = &4 k'&6 !s)'2
3 )4 N
7-/. 5t is deter!ined deter!ined that that a res#$tant res#$tant %orce o% o% 6 N wi$$ ie ie a waon an acce$erat acce$eration ion o% 1 !s). *hat %orce is re#ired to ie the waon an acce$eration o% on$y ) !s)+ 6 N
m=
)
1 ! s
= 6 s$#s 2
F = ma = &6 s$#s'&) !s)'2
3 1) N
7-6. A 1-k car !oin north north at 1 k!h "rakes to a stop in / !. *hat are the !anit#de and direction o% the %orce+ )as = v
) f
−v
) o
2
a=
Convert to SI units units 100 k!h = k!h = 27.8 ! s s v f)
− vo) ) s
=
&' )
− & )7.( ! s ' ) )&/ !'
F = ma = &1 k'&7.7) !s)'2
6/
2
a = 7.7) 7.7) ! s)
77) N, South. F
Physics, 6th Edition
Chapter 7 Newton’s Second Law The Relationship Between Weight and Mass
7-7. *hat is the weiht o% a 4.( k !ai$"o8+ *hat is the !ass o% a 4-N tank+ W = &4.( k'&9.( !s)' 47. N 2
m=
4 N 9.( ! s)
4.( k
7-(. *hat is the !ass o% a 6-$" chi$d+ *hat is the weiht o% a 7-s$# !an+ m=
6 $" )
) %t s
1.(( s$#s 2
W = &7 s$#s'&) %ts)' ))4 $"
7-9. A wo!an weihs 1( $" on earth. *hen she wa$ks on the !oon, she weihs on$y $". *hat is the acce$eration d#e to raity on the !oon and what is her !ass on the !oon+ :n the Earth+ Her mass is the same on the moon as it is on the earth, so we first find the constant mass me
=
1( $" )
) %t s
W m = mm $ m
. s$#s2 = /6)/
$ m
=
mm = me = !."2 s#u$s %
$" /.6)/ s$#s
%
$ m = !.&& %ts)
7-1. *hat is the weiht o% a 7-k astrona#t on the s#r%ace o% the earth. Co!pare the res#$tant %orce re#ired to ie hi! or her an acce$eration o% 4 !s) on the earth with the res#$tant %orce re#ired to ie the sa!e acce$eration in space where raity is ne$ii"$e+ 'n earth W = &7 k'&9.( !s)' 6(6 N 2 (n)where
F * = 280 N
3; &7 k'&4 !s)' )( N
+he mass doesnt chan$e.
7-11. 3ind the !ass and the weiht o% a "ody i% a res#$tant %orce o% 16 N wi$$ ie it an acce$eration o% / !s). 16 N m
=
)
/. ! , s
.) k
2
W = &.) k'&9.( !s)' 1.4 N
66
Physics, 6th Edition
Chapter 7 Newton’s Second Law
7-1). 3ind the !ass and weiht o% a "ody i% a res#$tant %orce o% ) $" ca#ses its speed to increase %ro! ) %ts to 6 %ts in a ti!e o% / s. a=
6 %t s - ) %t s /s
= ( %t s ) 2
m=
W = m$ &)/. s$#s'&) %ts)'2
) $" ( %t s)
)/. s$#s
* ( $"
7-1. 3ind the !ass and weiht o% a "ody i% a res#$tant %orce o% 4 N ca#ses it to decrease its e$ocity "y 4 !s in s. a=
∆v t
=
−4 ! s
2 a = −1 . ! s) 2
s
m=
W = m$ = & k'&9.( !s)'2
−4 N −1. ! s)
2
m = &00 k
W )94 N
Applications for Single-Bod !ro"le#s$
7-14. *hat hori
%riction %orce o% ) N opposes the !otion+ - = ) N &6 k'&4 !s)'2
) N
!
- 44. N
7-1/. A )/-$" a#to!o"i$e is speedin at // !ih. *hat res#$tant %orce is re#ired to stop the car in ) %t on a $ee$ road. *hat !#st "e the coe%%icient o% kinetic %riction+ We first find the mass and then the acce#eration. &// !ih (.7 !s' m=
a=
)/ $" )
) %t s v f )
− v) ) s
= 7(.1 s$#s2
=
Now reca$$ that )as = v f ) − v)
& ' − &(.7 %t s' ) )&) %t'
2
and
F = ma = &7(.1 s$#s'&-16. %ts)'2 F
= µ / 2
µ k =
1)7 $" )/ $"
67
2
a
=
- 16. ! s)
3 -1)7 $"
µk ./(
Physics, 6th Edition
Chapter 7 Newton’s Second Law
7-16. A 1-k !ass is $i%ted #pward "y a $iht ca"$e. *hat is the tension in the ca"$e i% the acce$eration is &a'
? T
/ote that u is ositive and that W = &1 k'&9.( !s)' 9( N. 2
a + 3 48 N &1 k'& !s ) 2
5 + 3 48 N &1 k'&6 !s ) 2
c + 3 48 N &1 k'&-6 !s )
1 k
W % mg
and
> 9( N
and
> 6 N ? 9( N
and
> - 6 N ? 9( N or
or
> 1/( N > (. N
7-17. A 64-$" $oad hans at the end o% a rope. 3ind the acce$eration o% the $oad i% the tension in ?
the ca"$e is &a' 64 $", &"' 4 $", and &c' 96 $". &a' + − W
=
W $
T
64 $" a 2 ) %ts)
a2 64 $" − 64 $"
&"' +
−W =
W
&"' +
−W =
W
$
$
m = W$
a=0
W
64 $" a 2 ) %ts)
a = 612.0 %ts)
64 $" a 2 ) %ts)
a = 1".0 %ts)
a2 4 $" − 64 $"
a2 96 $" − 64 $"
7-1(. An (-k e$eator is $i%ted ertica$$y "y a stron rope. 3ind the acce$eration o% the e$eator i% the rope tension is &a' 9 N, &"' 7(4 N, and &c' ) N.
T
/ewtons #aw for the ro5#em is + 3 m$ = ma u is ositive &a' 9 N = &( k'&9.( !s)' &( k'a % &a' 7(4 N = &( k'&9.( !s)' &( k'a % &a' ) N = &( k'&9.( !s)' &( k'a %
m
a = 1.4/ !s)
m$
a=0 a = -7. !s)
7-19. A hori
o% the ca"inet i% µk % .). F = µ N = µ k m$
1 N = F ma%
1 N
F 2
F = 0.2&( k'&9.( !s
) = 15.7 Ν
100 N = 1/.7 N &( k' a%
6(
a = 1./ !s)
m$
Physics, 6th Edition
Chapter 7 Newton’s Second Law 7-). 5n 3i. 7-1, an #nknown !ass s$ides down the inc$ined p$ane.
N
*hat is the acce$eration in the a"sence o% %riction+
Σ38 !a 9%
m$ sin &00 = ma % )
a = &9.( !s ' sin
a = $ sin &00 m$
)
4.9 !s , down the p$ane
7-)1. Ass#!e that µk % .) in 3i 7-1. *hat is the acce$eration+
N F
*hy did yo# not need to know the !ass o% the "$ock+
Σ38 !a 9%
m$ sin - µk N ma %
N = m$ cos
m$
m$ sin - µk m$ cos ma % a = $ sin - µk cos a = &9.( !s)'&./' = .)&9.( !s)'&.(66'2
a .) !s), down the p$ane.
@7-)). Ass#!e that m 5 k and µk . in 3i. 7- 1. *hat p#sh - directed #p and a$on the inc$ine in 3i.7-1 wi$$ prod#ce an acce$eration o% 4 !s) a$so #p the inc$ine+ F
µk N µk m$ cos 2 Σ38 !a2
F
N
.&1 k'&9.( !s)'cos )/./ N ?
- = F = m$ sin ma
F
- 3 2!.! N = &1 k'&9.( !s)'&./' &1 k'&4 !s)' - 3 2!.! N = 49. N 4 N2
!
m$
- 114 N
@7-). *hat %orce P down the inc$ine in 3i. 7-1 is re#ired to ca#se the acce$eration :*N the p$ane to "e 4 !s)+ Ass#!e that in 5: k and µk . . See -ro5. 7622 F is u the #ane now. - is down #ane . Σ F 9 = ma2 - - F ? m$ sin ma 2
Sti##, F )/./ N
- 6 2!.! N ? &1 k'&9.( !s)'&./' &1 k'&4 !s)' - 6 2!.! N ? 49. N 4 N2
- 16./ N
69
N
? !
F
m$
Physics, 6th Edition
Chapter 7 Newton’s Second Law Applications for M<i-Bod !ro"le#s
7-)4. Ass#!e in the connectin cord+
6 k
) k
*esu#tant force = tota# mass 9 acce#eration 80 N &) k ? 6 k'a%
a = 10 !s)
+o find +, a#) F = ma to "6$ 5#oc on#) 80 N = > &6 k'&1 !s
2
);
4/ N
N
T
connectin cord %or the arrane!ent shown in 3i. 7-1+ (ssume ;ero friction and draw free65od) dia$rams. For tota# s)stem m2 $ = m1 m2 a
=
B
3 1/ N
@7-)6. *hat are the acce$eration o% the syste! and the tension in the
a
/ k
A
a = & !s)
Force '/ : = m : a = &/ k'& !s)'2
Τ = 20 Ν
1 k
7-)/. *hat %orce does "$ock A e8ert on "$ock B in 3i, 7-1)+ F = m+ a% 4/ N &1/ k' a% Σ
( N
>
a T
m1 g
m1 $ is 5a#anced 5) N m2 g
)
m) $ m1 + m)
=
&6 k'&9.( ! s ' 4 k ? 6 k
%
a = !.88 !s) 2
F = m1a Σ
+ = m1a = &4 k'&/.(( !s
);
/ow, to find +, consider on#) m1
Τ = 23.5 Ν
@7-)7. 5% the coe%%icient o% kinetic %riction "etween the ta"$e and the 4 k "$ock is .) in
3i.
7-1, what is the acce$eration o% the syste!. *hat is the tension in the cord+ F ) = 0% Σ
N = m1
%$F = µ N = µ m1 $
F
N
'
T
a
For tota# s)stem m2 $ 6 µ m1 $ = m1 m2 a m) $ − µ m1 $ &6 k'&9.( ! s ' − & .)'&4 k'&9.( ! s ' )
a
=
m1 + m)
=
)
m1 g
T
4 k ? 6 k m2 g 7
Physics, 6th Edition
Chapter 7 Newton’s Second Law
@7-)7. &Cont.'
a=
/(.( N − 7.(4 N 1 k
a = /.1 !s)
or
'
+o find +, consider on#) m2 and mae down ositive F ) = m2a % Σ
m2 $ 3 + = m2a%
N
F
T
a T
m1 g
+ = m2 $ 3 m2a
)
)
Τ = &6 k'&9.( !s ' = &6 k'&/.1 !s '2
m2 g
> )(.) N
@7-)(. Ass#!e that the !asses m1 = ) k and m2 = ( k are connected "y a cord
a
that passes oer a $iht %riction$ess p#$$ey as in 3i. 7-14. *hat is the acce$eration o% the syste! and the tension in the cord+ T
T
*esu#tant force = tota# mass of s)stem 9 acce#eration a
m2 $ 3 m1 $ = m1 m2 a
=
m) $ − m1 $ m1 + m)
&( k'&9.( ! s) ' - &) k'&9.( ! s) '
a=
) k ? ( k
m2 g
m1 g
2
a = /.(( !s
/ow #oo at m1
a#one + 6 m1 $ = m1 a%
+ = m1$ a = &) k'&9.( !s) = /.(( !s)'2
> 1.4 N a
@7-)9. >he syste! descri"ed in 3i. 7-1/ starts %ro! rest. *hat is the N
T
acce$eration ass#!in
/1.9 N - 19.6 N 1) k
)
m1 $ sin ) = m2 $ = m1 m2 a
&1 k'&9.( !s)'sin ) = &) k'&9.( !s)' &1 k ? ) k'a a=
T
)
m1 g
a = ).69 !s)
@7-. *hat is the acce$eration in 3i. 7-1/ as the 1-k "$ock !oes down the p$ane aainst %riction &µk .)'. (dd friction force F u #ane in fi$ure for revious ro5#em.
71
m2 g
Physics, 6th Edition
Chapter 7 Newton’s Second Law m1 $ sin ) = m2 $ 3 F = m1 m2 a %
N = m1 $ cos )
F ) = 0 % Σ
@7-. &Cont.' m1 $ sin ) = m2 $ 3 F = m1 m2 a %
F
µk N
= µ m1 $ cos )
m1 $ sin ) = m2 $ 3 µ m1 $ cos ) = m1 m2 a %
a = 1.1 !s)
@7-1 *hat is the tension in the cord %or Pro"$e! 7-+ (#) F = ma to mass m2 on#) + = m2$ a = &) k'&9.( !s) ? 1.1 !s)'2
+ 3 m2 $ = m2 a%
> )).) N
Challenge !ro"le#s 7-).
A )-$" e$eator is $i%ted ertica$$y with an acce$eration o% ( %ts).
T (a
3ind the !ini!#! "reakin strenth o% the ca"$e p#$$in the e$eator+
Σ3y ma% + 3 m$ = ma%
m=
W $
=
) $" ) %t s)
+ = m$ a%
2 m = 6)./ s$#s
+ = &6)./ s$#s'&) %ts) ? ( %ts)'2
) $"
> )/ $"
7-. A )-$" worker stands on weihin sca$es in the e$eator o% Pro"$e! 7-). *hat is the readin o% the sca$es as he is $i%ted at ( !s+
(a
+he sca#e readin$ wi## 5e e
m$ = ma%
N
= m$ a%
+ = &6)./ s$#s'&) %ts) ? ( %ts)'2
N
) $"
> )/ $"
7-4. A (-k $oad is acce$erated #pward with a cord whose "reakin strenth is ) N. *hat is the !a8i!#! acce$eration+ + ma9 = m$ = ma
a
=
+!a8 − m$ m
=
) N - &( k'&9.( ! s) ' ( k
a 1/.) !s)
7)
T % ) N (a
8 k m$
Physics, 6th Edition
Chapter 7 Newton’s Second Law
7-/. 3or r#""er tires on a concrete road µk .7. *hat is the hori
+he stoin$ distance is determined 5) the
acce#eration from a resu#tant friction force F = µ N, where N m$ F
-µk m$ = ma2
) a = 6 µ $ = 6 &.7'&9.( !s '2
2
v f )
s =
2 f
*eca## that 2as = vo 6 v %
− v) )a
=
a = 66.(6!s)
− &) ! s' ) )&-6.(6 ! s) '
%
s = )9.) k!
@7-6. S#ppose the 4 and 6-k !asses in 3i. 7-1 are switched so that the $arer !ass is on the ta"$e. *hat wo#$d "e the acce$eration and tension in the cord ne$ectin %riction+ For tota# s)stem m2 $ = m1 m2 %
N
m1 = 6 k2 m2 = 4 k
T
a
=
a
)
m) $ m1 + m)
=
&4 k'&9.( ! s ' 6 k ? 4 k
%
a = .9) !s) 2
F = m1a Σ
T
m1 g
+ = m1a = &6 k'&/.(( !s
);
Τ = 23.5 Ν
m2 g
@7-7. Consider two !asses A and B connected "y a cord and h#n oer a sin$e p#$$ey. 5% !ass A is twice that o% !ass B, what is the acce$eration o% the syste!+ m ( = 2m : %
a
If the #eft mass : is m, the ri$ht mass ( wi## 5e 2m.
2m$ 3 m$ = 2m ma
T
m$ = &ma
T
B
a
$
9.( ! s)
= =
A
2m
m 2
a = .)7 !s
2m g
m g
@7-(. A /-k !ass rests on a 4 inc$ined p$ane where µk .). *hat p#sh
N
?
#p the inc$ine, wi$$ ca#se the "$ock to acce$erate at 4 !s)+
!
F
4 F
µk N µk m$ cos 42
F
.)&/ k'&9.( !s)'cos 4 (.1) N 4
Σ38 !a2
- = F = m$ sin 4 ma
- 3 8.12 N = &/ k'&9.( !s)' sin 4 &/ k'&4 !s)'
7
- 47.4 N
m$
Physics, 6th Edition
Chapter 7 Newton’s Second Law
@7-9. A 96-$" "$ock rests on a ta"$e where µk % .4. A cord tied to this "$ock passes oer a $iht %riction$ess p#$$ey. *hat weiht !#st "e attached to the %ree end i% the syste! is to acce$erated at 4 %ts)+ F
N
µk N .) &96 $"'2
* - 19.) $"
F
T
F
19.) $"
96 $" ? * 4 %ts) ) ) %ts ) (
* = 19.) $" 1) $" ? .1)/ *2
a T
96 $" W
* /.7 $"
Critical Thin'ing )&estions
7-4. 5n a $a"oratory e8peri!ent, the acce$eration o% a s!a$$ car is !eas#red "y the separation o% spots "#rned at re#$ar intera$s in a para%%in-coated tape. Larer and $arer weihts are trans%erred %ro! the car to a haner at the end o% a tape that passes oer a $iht %riction$ess p#$$ey. 5n this !anner, the !ass o% the entire syste! is kept constant. Since the car !oes on a horihe %o$$owin data are recorded *eiht, N Acce$eration, !s)
)
4
6
(
1
1)
1.4
).9
4.1
/.6
7.1
(.4
P$ot a raph o% weiht &%orce' ers#s acce$eration. *hat is the sini%icance o% the s$ope o% this c#re+ *hat is the !ass+
14 12
+he s#oe is the chan$e in Force over the chan$e in acce#eration, which is the
N10 , e 8 c r o 6 F
∆ F
4 2
mass of the s)stem. +hus, the mass is found to 5e
∆a
0 0
m = 1.4) k
1
2
3
4
5
6
Acce$eration, !s
74
)
7
8
9
Physics, 6th Edition
Chapter 7 Newton’s Second Law
7-41. 5n the a"oe e8peri!ent, a st#dent p$aces a constant weiht o% 4 N at the %ree end o% the tape. Seera$ r#ns are !ade, increasin the !ass o% the car each ti!e "y addin weihts. *hat happens to the acce$eration as the !ass o% the syste! is increased+ *hat sho#$d the a$#e o% the prod#ct o% !ass and acce$eration "e %or each r#n+ 5s it necessary to inc$#de the !ass o% the constant 4 N weiht in these e8peri!ents+ +he acce#eration increases with increasin$ mass. (ccordin$ to /ewtons second #aw, the roduct of the tota# mass of the s)stem and the acce#eration must a#wa)s 5e e
7-4). An arrane!ent si!i$ar to that descri"ed "y 3i. 7-1 is set #p e8cept that the !asses are rep$aced. *hat is the acce$eration o% the syste! i% the s#spended
N
F
'
T
a
!ass is three ti!es that o% the !ass on the ta"$e and µk .. Σ F ) = 0%
N = m
%$F = µ N = µ m$
T
mg
&mg
For tota# s)stem &m$ 6 µ m$ = &m ma % & 6 µ m$ = ma a=
& − µ ' $ 4
=
& - .'&9.( ! s) '
a = 6.6) !s)
4
7-4. >hree !asses, ) k, 4 k, and 6 k, are connected &in order' "y strins and h#n %ro! the cei$in with another strin so that the $arest !ass is in the $owest position. *hat is the tension in each cord+ 5% they are then detached %ro! the cei$in, what !#st "e the tension in the top strin in order that the syste!
A ) k
B 4 k
C
acce$erate #pward at 4 !s)+ 5n the $atter case what are the tensions in the strins that connect !asses+ 7-4. &Cont.' +he tension in each strin$ is due on#) to the wei$hts :>?'W the strin$. +hus, 7/
6 k
Physics, 6th Edition
Chapter 7 Newton’s Second Law + C = &6 k'&9.( !s)' /(.( N 2
+ : = &6 k ? 4 k'&9.( !s)' 9(. N 2
+ ( = &6 k ? 4 k ? ) k'&9.( !s)' 2
/ow consider the uward acce#eration of 4 !s
Σ
11( N
.
3y 2 + ( &) k ? 4 k ? 6 k'&4 !s)'2
A
a = ms 2
) k
B
>A 4( N
4 k
C 6 k
+ : = &4 k ? 6 k'&4 !s)' 4 N 2 >C &6 k'&4 !s)' )4 N
7-44. An (-k astrona#t on a space wa$k p#shes aainst a )-k so$ar pane$ that has "eco!e dis$oded %ro! a spacecra%t. >he %orce ca#ses the pane$ to acce$erate at ) !s). *hat acce$eration does the astrona#t receie+ o they contin#e to acce$erate a%ter the p#sh+ +he force on the so#ar ane# F is e
=−
m a ma
=−
+hus,
m a = 6 maaa % so#ve for aa
&) k'&) ! s) ' ( k
2
a = - / !s)
(cce#eration e9ists on#) whi#e a force is a#ied, once the force is removed, 5oth astronaut and so#ar ane# move in oosite directions at the seeds o5tained when contact is 5roen..
7-4/. A 4-$" s$ed s$ides down a hi$$ &µk .)' inc$ined at an an$e o% 6. *hat is the nor!a$ %orce on the s$ed+ *hat is the %orce o% kinetic %riction+ *hat is the res#$tant %orce down the hi$$+ *hat is the acce$eration+ 5s it necessary to know the weiht o% the s$ed to deter!ine its N
acce$eration+
Σ3y 0%
N =
* cos 6 2 N &4 $"'cos 6
F
µk N &.)'&) $"'2
F
) $" 2
4 $"
Σ38 W sin 6 = F = 00 $"'sin 6 = 4 $"2
76
F
6
?a 6
3; 6 $"
* ! 4 $"
Physics, 6th Edition
Chapter 7 Newton’s Second Law
7-4/ &Cont. Since F * ma% we note that W sin "00 6 µ W = W$a% +hus, the wei$ht divides out and it is not necessar) for determinin$ the resu#tant acce#eration.
@7-46. >hree !asses, m1 = 1 k, m2 = ( k, and m& = 6 k, are connected as shown in 3i. 7-16. Ne$ectin %riction, what is the acce$eration o% the syste!+ *hat are the tensions in the cord on the $e%t and in the cord on the riht+ *o#$d the acce$eration "e the sa!e i% the N
!idd$e !ass m2 were re!oed+
+ :
+ (
?
?
+ota# mass of s)stem = 10 8 " = )4 k + (
*esu#tant Force on s)stem = m1 $ 3 m& $ +he norma# force N 5a#ances m2 $% m1 $ 3 m& $ = m1 m2 m& a %
+ :
m2 $
1 k
6 k
( k
m& $
m1 $
Σ3 m+ a
&1 k'&9.( !s)' = &6 k'&9.( !s)' &)4 k' a a = 1.6 !s) 2 +he acce#eration is not affected 5) m2.
&)4 k'a 9(. N = /(.( N2
+o find + ( a#) Σ F = m1a to 16k !ass m1 $ 3 + ( = m1a % 2
+ ( = m1$ 3 a = &1 k'&9.( !s /ow a#) to " -k !ass 2
+ : = &6 k'&9.( !s
+
−
1.6 !s)'2
+ : 3 m& $ = m&a%
1.6 !s)' %
+ ( = m1 $ 3 m1a
+ ( = (1.7 N
+ : = m& $ m&a
+ : 6(.6 N
@7-47. Ass#!e that µk % . "etween the !ass m2 and the ta"$e in 3i. 7-16. >he !asses m2 and m& are ( and 6 k, respectie$y. *hat !ass m1 is
N
?
F
+ :
+ (
?
re#ired to ca#se the syste! to acce$erate to the $e%t at ) !s)+
& F µk m2 $ acts to ri$ht.
(#) Σ F = m+ a to tota# s)stem, #eft is ositive.
+ ( m1 $
m1 $ 3 F 3 m& $ = m1 m2 m& a % F µk m2 $ = 0.&&( k'&9.( !s)'2
77
+ :
m2 $ ( k
F
)./ N
6 k
m& $
Physics, 6th Edition
Chapter 7 Newton’s Second Law
m1&9.( !s)' = )./ N - &6 k'&9.( !s)' &m1 14 k'&) !s)' 9.( m1 3 )./ k = /(.( k )m1 28 k 2
m1 = 14.1 k
@7-4(. A "$ock o% #nknown !ass is ien a p#sh #p a 4 inc$ined p$ane and then re$eased. 5t contin#es to !oe #p the p$ane &?' at an acce$eration o% =9 !s). *hat is the coe%%icient o% kinetic %riction+
F
4
Since 5#oc is movin$ u #ane, F is directed down #ane. F
µk N 2 Σ3y 2
Σ38 !a2
N
4
m$
! cos 42 F µk ! cos 4
-F - m$ sin 4 ma2
a = 6 µ $ sin 4 2 cos 4 - $
F
- µ m$ cos 4 - m$ sin 4 ma
-9 !s) -µk &9.( !s)' cos 4 - &9.( !s)' sin 4
So#vin$ for µ we o5tain
µ = .6
@7-49. B$ock A in 3i. 7-17 has a weiht o% 64 $". *hat !#st "e the weiht o% "$ock B i% B$ock A !oes #p the p$ane with an acce$eration o% 6 %ts). Ne$ect %riction.
Σ38 ma
*B = *A sin 6 &m ( m : a
?a
N
W :
W + W − W( sin 6 = ( : a % $
a
o
$
=
6 %ts
)
) %ts )
T
= .1((
T 6 6
W : 3 &64 $"'&.(66' .1((&64 $" ? *B' W : 3 //.4 $" 1). $" ? .1((*B 2
*B
*A 64 $"
*B (. $"
@7-/. >he !ass o% "$ock B in 3i. 7-17 is 4 k. *hat !#st "e the !ass o% "$ock A i% it is to !oe down the p$ane at an acce$eration o% ) !s)+ Ne$ect %riction.
Σ
38 ma
T
m ( $ sin 6 - mB &m ( m : a
)
)
)
T
(.49 m ( = 9.) k ) m ( ? ( k2
m ( 7.)( k.
7(
6
)
&9.( !s '&.(66'm ( 3 &4 k'&9.( !s ' m ( ) !s ' ? &4 k'&) !s '
?a
N
6
m ( $
m : $
Physics, 6th Edition
Chapter 7 Newton’s Second Law
@7-/1. Ass#!e that the !asses A and B in 3i. 7-17 are 4 k and 1 k, respectie$y. >he coe%%icient o% kinetic %riction is .. 3ind the acce$eration i% &a' the syste! is initia$$y !oin #p the p$ane, and &"' i% the syste! is initia$$y !oin down the p$ane+ &a' With uward initia# motion, F is down the #ane. F
µk N 2 Σ3y 2
N m ( $ cos
?a
N
T
62 F µk m ( $ cos 6
T F
*esu#tant force on entire s)stem = tota# mass 9 acce#eration m : $ 3 m ( $ sin 6 - µk m ( $ cos 6 &m ( m :'a
6 6
m : $
m ( $
&1 k'&9.( !s)' = &4 k'&9.( !s)'&.(66' = .&1 k'&9.( !s)'&./' &14 k'a 9( N = .9 N = 14.7 N &14 k'a%
or
a ./ !s)
&"' If initia# motion is down the #ane, then F is u the #ane, 5ut the resu#tant force is sti## down the #ane. +he 5#oc wi## side unti# it stos and then $oes the other wa). Σ F 9 = ma%
m : $ 3 m ( $ sin 6 ? µk m ( $ cos 6 &m ( m :'a 9( N = .9 N ? 14.7 N &14 k'a
a /.6 !s)
?a
N
T T
F
6
+he $reater acce#eration resu#ts from 6
the fact that the friction force is increasin$ the resu#tant force instead of decreasin$ it as was the case in art a.
79
m ( $
m : $
