Discussion Class Solutions
Week 8: 17 – 21 March
HRW 9th ed. Ch 6: P18, P25, P27
HRW 9th ed. Ch 7: Q1, Q4, Q5, Q6, Q7, Q9, P5, P9, P21, P27
Chapter 6
P18.
a) We apply Newton’s second law to the “downhill” direction:
mg sin – f = ma,
where, using Eq. 6-11,
f = fk =kFN =k mg cos
Thus, with k = 0.600, we have
a = gsin – kg cos = –3.72 m/s2
which means, since we have chosen the positive direction in the direction of motion (down the
slope) then the acceleration vector points “uphill”; it is decelerating. With v0  18.0 m/s and x = d
= 30.0 m, Eq. 2-16 leads to
v  v02  2ad  10.0 m/s.
b) In this case, we find a = +1.1 m/s2, and the speed (when impact occurs) is 19.7 m/s.
P25.
The free-body diagrams for block B and for the knot just above block A are shown below.


T1 is the tension force of the rope pulling on block B or pulling on the knot (as the case may be), T2

is the tension force exerted by the second rope (at angle  = 30°) on the knot, f is the force of
static friction exerted by the horizontal surface on block B, FN is normal force exerted by the

surface on block B, WA is the weight of block A (WA is the magnitude of m A g ), and WB is the weight

of block B (WB = 750 N is the magnitude of mB g ).
For each object we take +x horizontally rightward and +y upward. Applying Newton’s second law in
the x and y directions for block B and then doing the same for the knot results in four equations:
T1  f s ,max
FN  WB
T2 cos   T1
T2 sin   WA




0
0
0
0
where we assume the static friction to be at its maximum value (permitting us to use Eq. 6-1).
Solving these equations with s = 0.25, we obtain WA  108 N  1.0 10 2 N .
P27.
First, we check to see if the bodies start to move. We assume they remain at rest and compute the
force of (static) friction which holds them there, and compare its magnitude with the maximum
value sFN. The free-body diagrams are shown below.
T is the magnitude of the tension force of the string, f is the magnitude of the force of friction on

body A, FN is the magnitude of the normal force of the plane on body A, m A g is the force of gravity

on body A (with magnitude WA = 102 N), and mB g is the force of gravity on body B (with magnitude

WB = 32 N).  = 40° is the angle of incline. We are told the direction of f but we assume it is
downhill. If we obtain a negative result for f, then we know the force is actually up the plane.
a) For A we take the +x to be uphill and +y to be in the direction of the normal force. The x and y
components of Newton’s second law become
T  f  WA sin   0
FN  WA cos   0.
Taking the positive direction to be downward for body B, Newton’s second law leads to
WB  T  0 . Solving these three equations leads to
f  WB WA sin   32 N  (102 N)sin 40  34 N
(indicating that the force of friction is uphill) and to
FN  WA cos   (102 N) cos 40  78N
which means that
fs,max = sFN = (0.56) (78 N) = 44 N.
Since the magnitude f of the force of friction that holds the bodies motionless is less than fs,max the
bodies remain at rest. The acceleration is zero.
b) Since A is moving up the incline, the force of friction is downhill with magnitude f k  k FN .
Newton’s second law, using the same coordinates as in part (a), leads to
T  f k  WA sin   mA a
FN  WA cos   0
WB  T  mB a
for the two bodies. We solve for the acceleration:
a
32N  102N  sin 40   0.25 102N  cos 40
WB  WA sin    kWA cos 

mB  mA
 32N +102N   9.8 m s 2 
  3.9 m s 2 .
The acceleration is down the plane, i.e., a  ( 3.9 m/s 2 )iˆ , which is to say (since the initial velocity
was uphill) that the objects are slowing down. We note that m = W/g has been used to calculate the
masses in the calculation above.
c) Now body A is initially moving down the plane, so the force of friction is uphill with magnitude
f k  k FN . The force equations become
T  f k  WA sin   mA a
FN  WA cos   0
WB  T  mB a
which we solve to obtain
a
32N  102N  sin 40   0.25 102N  cos 40
WB  WA sin    kWA cos 

mB  mA
 32N +102N   9.8 m s 2 
 1.0 m s 2 .
The acceleration is again downhill the plane, i.e., a  ( 1.0 m/s 2 ) ˆi . In this case, the objects are
speeding up.
Chapter 7
Q1.
All the same. All cases have v=5m.s-1 thus all cases have kinetic energy that is the same since the
mass in each case is the same.
Q4.
C, B, A. Since each case has the same mass one needs only look at 𝑣𝑓2 − 𝑣𝑖2 in order to rank the the
change in kinetic energy. In A this value is -11, in B it is -7 and in C it is +12.
Q5.
B, A, C, D. Work is area under the curve of a F vs x graph.Positive work occurs when the F>0, while
negative work occurs when F<0.
Q6.
(a) Greatest speed and kinetic energy is at x=3. Since 𝑊 = ∫ 𝐹 𝑑𝑥 = area under curve and 𝑊𝑛𝑒𝑡 =
∆𝐸𝐾 then the place where the greatest speed/kinetic energy is the place where the area under the
curve is at its largest positive value. The particle has zero speed again at x=6 since then area above
the curve is the same as the area below the curve thus W=0 and hence initial kinetic energy is the
same as the final kinetic energy hence 𝑣𝑓 = 𝑣𝑖 = 0
(b) 𝑊𝑛𝑒𝑡 < 0 at this stage hence ∆𝐸𝐾 < 0 thus since particle started off at rest then direction of
motion must be in negative x direction. (i.e. it starts off stationary and gains kinetic energy thus
moves off in positive direction due to positive forces acceleration, and then starts to lose kinetic
energy after 3 and slows down until it at rest again at 6, after which the particle still experiences
negative acceleration and starts moving in the negative direction)
Q7.
All the same. Along each path the pig descends the same vertical distance hence work done by
gravity is the same in each case.
Q9.
1
1
(a) A will do more work. 𝑊 = 2 𝑘𝑥𝑖2 − 2 𝑘𝑥𝑓2 thus if they are compressed the same distance (𝑥𝑖2 −
𝑥𝑓2 is the same) then the spring with the higher spring constant does more work
(b) B will do more work. Since F=-kx then by keeping F the same in both cases this means that the
distance compressed for A will be proportionally smaller than the distance compressed for B e.g.
if A is twice as stiff as B then B will be compressed twice as much. However work relies on the
square of the compressed distance so then B will do more work since it has been compressed
1
the larger distance e.g. the work done by spring A in our previous case 𝑊𝐴 = − 2 𝑘𝐴 𝑥𝐴2 =
1
𝑥
2
1
− 2 2𝑘𝐵 ( 2𝐵 ) = − 4 𝑘𝐵 𝑥𝐵2 =
𝑊𝐵
2
P5.
We denote the mass of the father as m and his initial speed vi. The initial kinetic energy of the father
is
Ki 
1
Kson
2
and his final kinetic energy (when his speed is vf = vi + 1.0 m/s) is K f  Kson . We use these relations
along with Eq. 7-1 in our solution.
(a) We see from the above that Ki  21 K f , which (with SI units understood) leads to
1 2 1 1
2
mvi 
m  vi  1.0 m/s   .

2
2 2

The mass cancels and we find a second-degree equation for vi :
1 2
1
vi  vi   0.
2
2
The positive root (from the quadratic formula) yields vi = 2.4 m/s.
(b) From the first relation one gets
1 2 1 1
2 
mvi   (m/2) vson

2
2 2

and (after canceling m and one factor of 1/2) are led to vson = 2vi = 4.8 m s.
P9.
By the work-kinetic energy theorem,
W  K 


1 2 1 2 1
mv f  mvi  (2.0 kg) (6.0 m/s) 2  (4.0 m/s) 2  20 J.
2
2
2


We note that the directions of v f and v i play no role in the calculation.
P21.
We use F to denote the magnitude of the force of the cord on the block. This force is upward,
opposite to the force of gravity (which has magnitude Fg  Mg ), to prevent the block from

undergoing free fall. The acceleration is a  g / 4 downward. Taking the downward direction to be
positive, then Newton’s second law yields
𝐹⃗𝑛𝑒𝑡 = 𝑚𝑎⃗ ⇒ 𝑀𝑔 − 𝐹 = 𝑀
𝑔
4
so F = 3Mg/4, in the opposite direction of the displacement. On the other hand, the force of gravity
Fg  mg is in the same direction to the displacement.
(a) Since the displacement is downward, the work done by the cord’s force is, using Eq. 7-7,
3
WF   Fd   Mgd .
4
(b) Similarly, the work done by the force of gravity is Wg  Fg d  Mgd .
(c) The total work done on the block is simply the sum of the two works:
3
1
Wnet  WF  Wg   Mgd  Mgd  Mgd .
4
4
Since the block starts from rest, we use Eq. 7-15 to conclude that this is block’s kinetic energy K at
the moment it has descended the distance d.
(d) Since K  12 Mv 2 , the speed is
v
2K
2( Mgd / 4)


M
M
gd
2
at the moment the block has descended the distance d.
P27.
From Eq. 7-25, we see that the work done by the spring force is given by
Ws 
1
k ( xi2  x 2f ) .
2
The fact that 360 N of force must be applied to pull the block to x = + 4.0 cm implies that the spring
constant is
360 N
 90 N/cm  9.0 103 N/m .
4.0 cm
(a) When the block moves from xi   5.0 cm to x  3.0 cm , we have
k
1
(9.0 103 N/m)[(0.050 m) 2  (0.030 m) 2 ]  7.2 J.
2
(b) Moving from xi   5.0 cm to x  3.0 cm , we have
Ws 
Ws 
1
(9.0 103 N/m)[(0.050 m) 2  ( 0.030 m) 2 ]  7.2 J.
2
(c) Moving from xi   5.0 cm to x  5.0 cm , we have
Ws 
1
(9.0 103 N/m)[(0.050 m) 2  (0.050 m) 2 ]  0 J.
2
(d) Moving from xi   5.0 cm to x  9.0 cm , we have
Ws 
1
(9.0 103 N/m)[(0.050 m) 2  ( 0.090 m) 2 ]  25 J.
2