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Forces and Laws of Motion
Worksheet 4.6 –Applications of Newton’s Laws
For each of the problems below, you must begin your solution with a free body diagram
1. A 20.0 kg crate hangs at the end of a long rope. Find the size and direction of its acceleration when
the tension in the rope is
a) 250 N
b) 150 N
FT=250N
a
FT=150N
a
Fg=mg=196N
Fy  ma y
FT  Fg  ma
250  196  20a
a  2.7 m / s 2 up
Fg=mg=196N
Fy  ma y
FT  Fg  ma
150  196  20a
a  2.3m / s 2 down
2. In a modified tug of war game, two people pull in opposite directions, not on a rope, but on a 25.0
kg sled resting on an icy road. If the participants exert forces of 90.0 and 92.0 N, how much will
the sled accelerate?
Fx  max
92  90  25a
a  0.08m / s 2 in same direction as 92 N force
3. Compute the size of the initial upward acceleration of a rocket of mass 1.30 x 104 kg if the initial
upward thrust of its engine is 2.60 x 105 N. Do not neglect the weight of the rocket. Construct the
force diagram
Fthrust
=260,000 N
Fy  ma y
FThrust  Fg  ma y
260,000  127,400  13,000a
a  10.2m / s 2 up
a
Fg=mg=127,400N
4. A 60 kg woman stands in an elevator that is accelerating downward at 3.10 m/s2.
a) Construct the force diagram for the woman
b) What is her weight?
FN
Fg  mg  588N
c) What is her apparent weight? (measured by the normal force
acting on her)
a=3.10
F  ma
y
FN  Fg
FN
FN
FN
y
 ma y
 Fg  ma
 588  60(3.1)
 402N
Fg=mg=588N
d) The elevator now accelerates up at 3.10 m/s2. Construct the force diagram
and determine the woman’s apparent weight.
Fy
FN  Fg
FN
FN
FN
 ma y
 ma y
 Fg  ma
 588  60(3.1)
 774N
FN
a=3.10
Fg=mg=588N
5. The maximum force that a grocery bag can withstand without ripping is 250 N. Suppose that bag is
filled with 20.0 kg of groceries and lifted with an acceleration of 5.0 m/s2. Do the groceries stay in
the bag?
Flift
Fy
Flift  Fg
Flift
Flift
Flift
 ma y
 ma y
 Fg  ma
 196  20(5)
 296 N
a
The bag rips
Fg=mg=196N
6. Suppose that a 1000 kg car is traveling at 25 m/s (55 mph). Its brakes can apply a force of 5000N.
What is the minimum stopping distance required for the car to stop?
v0x=25 m/s
vfx=0
Fbrake=5000N
Fx  max
 Fbrake  ma y
 5000  1000a
a  5m / s 2
Dx=?
v 2fx  v02x  2aDx
0  252  2(5)(Dx)
Dx  62.5m
7. Below is a picture of an Atwood’s machine: two masses attached to a frictionless, massless pulley.
The mass of block A is 5.0 kg and the mass of block B is 2.0 kg.
a) What is the acceleration of the system when the
blocks are released?
See the lecture powerpoint for how to solve this
problem by applying Newton’s 2nd Law to each
block and using the 2 equations to solve for the two
unknowns (FT and a).
FT
FT
a
A
B
FgA = mg = 49N
a
Here I will show how to solve the problem as a
system. If we consider block A and B as a system
of mass mA + mB, then the force of gravity pulling
block A down, FgA, is opposed only by the tension,
FT which is the force of gravity pulling on B, FgB,
since FT’s are internal forces that cancel to 0.
Newton’s 2nd Law for the SYSTEM:
FgB = mg = 19.6N
FgB = mg = 19.6N
B
Fsys y  msys a y
 FgA  FgB  (m A  mB )a y
 49  19.6  (5  2)a y
a y  4.2m / s 2
a
FT
FT
A
FgA = mg = 49N
Internal
forces
External
forces
b) How long will it take for block A to fall 2.0 m?
v0 y  0
v fy
a y  4.2m / s 2
Dy  2m
t
Dy  v0 y t  12 a y t 2
 2  12 (4.2)t 2
t  0.98s
8. A shuffleboard disk is accelerated to a speed of 5.8 m/s and released. If the coefficient of kinetic
friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to
a stop? The courts are 15.8 m long.
FN
fK
Vfx=0
V0x=5.8m/s
Dx=?
Fg
Need to find acceleration. Only force acting in the direction of motion(x) is friction.
Fx  max
 f k  max
  k FN  max
 (0.31)( Fg )  max
 (0.31)(mg )  max
a x  (0.31)(9.8)
a x  3.04m / s 2
v 2fx  v02x  2aDx
0  5.82  2(3.04)(Dx)
Dx  5.53m
9. Stacie, who has a mass of 45 kg, starts down a slide that is inclined at an angle of 45° with the
horizontal. If the coefficient of kinetic friction between Stacie’s shorts and the slide is 0.25, what is
her acceleration?
x-axis is parallel to the incline and y-axis is perpendicular
F
N
Fx  max
 f k  Fgx  max
  k FN  mg sin 45  max
  k Fgy  mg sin 45  max
  k mg cos 45  mg sin 45  max
 (0.25)(9.8)(0.707)  (9.8)(0.707)  a x
a x  5.2m / s 2
down the incline
fK
Fgy
45o
45o
Fgx
Fg=mg=441N
10. A 63-kg water skier is pulled up a 14.0° incline by a rope parallel to the incline with a tension of
512 N. The coefficient of kinetic friction is 0.27. What are the magnitude and direction of the
skier’s acceleration?
FN
x-axis is parallel to the incline and y-axis is perpendicular
Fx  max
 f k  Fgx  FT  max
  k FN  Fg sin 14  512  max
  k Fgy  Fg sin 14  512  max
  k Fg cos 14  Fg sin 14  512  max
 (0.27)(617.4)(0.970)  (617.4)(0.242)  512  (63)a x
a x  3.19m / s 2
14o
14o
Fgy
up the incline
Fg=617.4N
Fgx
11. A 15 kg block is on a ramp that is inclined at 20° above the horizontal. It is connected by a string
to a 19 kg mass that hangs over the top edge of the ramp. The coefficient of kinetic friction
between the incline and the 15 kg block is 0.22. What is the magnitude of the acceleration of the
19 kg block? (See the figure.)
FN
a(+)
FT
FT
a(+)
fk
Fgy
Fg1=mg
Fgx
Fg2=mg =186.2N
15 kg Block on incline:
Fx
 f K  Fgx  FT
  k FN  mg sin 20  FT
 (0.22) Fgy  (15)(9.8) sin 20  FT
 (0.22)(mg cos 20)  50.28  FT
 30.39  50.28  FT
FT
19 kg hanging block:
 max
 max
 15a
 15a
 15a
 15a
 15a  80.67
Fy  ma y
 FT  Fg  ma
 FT  186.2  19a
FT  19a  186.2
15a  80.67  19a  186.2
34a  105.53
a  3.1m / s 2 down
You can also solve the problem by treating the 2 blocks as one system since they are connected by one
rope and accelerate the same. Then the forces of tension are internal forces that cancel out within
the system. The force of gravity on the 19kg block is opposed by the forces parallel to the incline
(fK and Fgx)
For the system:
Fy  msys a y
a(+)
FN
Fg 2  f K  Fgx  (15  19)a y
186.2   k FN  mg sin 20  34a y
FT
FT
a(+)
fk
186.2  (0.22) Fgy  (15)(9.8) sin 20  34a y
Fgy
186.2  (0.22)(mg cos 20)  50.28  34a y
Fg1=mg
186.2  30.39  50.28  34a y
Fgx
a y  3.1m / s 2
12. In the figure to the right, two boxes of masses m=2kg and 4m=8 kg are in contact
with each other on a frictionless surface. A force of F=50
N is applied to m.
a) What is the acceleration of the more massive box?
Both boxes accelerate the same so treat both boxes as a
system:
Fg2=mg =186.2N
Fx  msys a x
Fpush  (5m)a
50  (2  8)a
a  5m / s 2
b) What is the force causing the acceleration of the more massive box (the net force on 4m)?
The force causing the acceleration of the larger box is the force applied from the smaller box,
F4m by m.
For the larger box:
Fx  max
F4 m by m  (8)(5)  40 N
F4m,m
F = 50N
m
4m
c) What is the force exerted on the smaller box by the larger box?
It is equal and opposite to the force exerted on the larger box by the smaller box: 40 N
For the smaller box:
Fx  max
F  Fm by 4 m  2a
50  Fm by 4 m  (2)(5)
Fm by 4 m  40 N
F = 50N
Fm,4m
m
4m
4m
d) What is the net force on the smaller box?
m
Fx  F  Fm by 4m  50  40  10 N
F = 50N
F4m,m
//
//
Fm,4m
\
e) Show all the action reaction pairs of forces for all the forces acting on the boxes.
Action-reaction force pairs are shown in the same color. For every action there is an equal and
opposite reaction. Remember action and reaction forces act on different objects and therefore do
not cancel out
FN (on 4m by table)
FN (on m by table)
F(on m by 4m)
F(on 4m by m)
FN (on table by m)
Fg (on m by E)
FN (on table by 4m)
Fg (on 4m by E)
Fg (on Earth by m)
Fg (Earth by 4m)
f) with friction (k = 0.2)
Both boxes accelerate the same so treat both boxes as a system:
Fx  msys a x
Fpush  f k  (5m)a
F   k FN
a  push
5m
50  0.2 Fg 50  0.2(10 g )


10
10
2
a  3.04m / s
FN
System
4m
m
F = 50N
fk
Fg
Net force on 4m box: Since its acceleration is 3.04m/s in +x direction:
Fx  Fnet  4max
Fnet  8(3.04)
 24.3N
Force exerted on the smaller box by the larger box?
Fx  m1a x
F12
F  F12  f k  max
fk
F12  F  f k  max
 F   k FN 1  max
 50  0.2(2)(9.8)  2(3.04)
F12  40 N
FN1
F = 50N
4m
m
Fg1
net force on the smaller box. Since its acceleration is 3.04m/s in +x direction:
Fx  Fnet  max
Fnet  2(3.04)
 6.08 N
13. This arrangement is being pulled across the
floor. The coefficient of kinetic friction
between the larger block and the floor is (0.50).
The coefficient of static friction between the
two blocks is (0.75). (Note that the weight, not
the mass, of each block is given)
a) What is the maximum acceleration of the blocks before the top block begins to slide off the
bottom block? (Hint: Draw the free body diagram of the top block and determine the
acceleration when the coefficient of static friction is at its max value of 0.75 and the top block
is at rest relative to the bottom block)
F12
For the blocks to move together, the top block is held in
place by static friction. Without friction between the
blocks, the top block would slide off the back of the
bottom block. Therefore, the force of static friction is
fs1
30N
opposite this motion, or pointing to the right. At right
is shown the FBD of the top block. The force on the
Fg1
top block by the bottom (F12) acts as a normal force.
Max acceleration of the blocks before top slides off is
100N
when fs1 is maximum. Net force on top block:
Fx
f s1
 s F12
 s m1 g
ax
 m1a x
 m1a x
 m1a x
 m1a x
  s g  7.35m / s 2
F
b) What is the maximum horizontal Force, F, that can be exerted on the lower block before the top
block begins to slide off? (Hint: Draw the free body diagram of the bottom block and apply
Newtons 2nd law to the bottom block)
The max force is the force that results in an acceleration of
7.35m/s2, found in part a).
FN
Note that F21=F12=Fg1 = 30N.
Net force on the bottom block:
Fx  m2 a x
F  f k 2  m2 a x
F   k FN  m2 a x
F   k ( Fg 2  F21 )  m2 a x
F  m2 a x   k ( Fg 2  F21 )
 10.2(7.35)  0.5(100  30)
F  140 N
30N
F
100N
fk2
F21
Fg2
Fy  0
FN  Fg 2  F21  0
FN  Fg 2  F21  100  30  130 N
(Normal force bears the weight of both boxes)
14. Two blocks of mass m1 = 2 kg and mass m2 = 7 kg are wedged up
against one another and against a wall by a horizontal force F. Doing
each section algebraically before putting in numbers:
a) What is the coefficient of static friction between m1 and m2 (call
this s1) and between m2 and the wall (call this s2) if the
MINIMUM force Fmin required to keep the blocks from breaking
loose and sliding under the influence of gravity if Fmin = 25 N?
m1
m2
F
To find the friction between the blocks and the friction between the block and wall, need to look at
each block separately. The freebody diagrams for each block are shown below. In each case, the
force balancing gravity and preventing the blocks from falling is static friction.
fs1
F12
//
m1
fs2
m2
F
//
wall
wall
m1
F
FN
//
m2
Fg1
Fg2
F21
//
wall
Block 1
Block 2
Fy  m1a y  0
Fy  m2 a y  0
 Fg1  f s1  0
 Fg 2  f s 2  0
 m1 g   s1 FN  0
 m2 g   s 2 FN  0
 s1 F12  m1 g
 s1 
 s 2 F  m2 g
m1 g m1 g

F12
F
m2 g
F
 2.74
s2 
s2
 s1  0.784
b) The force F is decreased to 20 N. The blocks break loose and begin to fall. If the coefficient of
kinetic friction between m1 and m2 AND between m2 and the wall are k1 = 0.15 and k2 = 0.9
respectively, what are the accelerations of m1 and m2? Note that they will not be the same.
fk1
F12
//
m1
m2
fk2
wall
F
//
wall
m1
F
FN
m2
//
F21
//
Fg1
Fg2
Block 1
Block 2
Fy  m1a1 y  0
Fy  m2 a2 y  0
 Fg1  f k1  m1a1 y
 Fg 2  f k 2  m2 a2 y
 m1 g   k1 FN  m1a1 y
 m2 g   k 2 FN  m2 a2 y
a1 y 

 m1 g   k1 F12
m1
 m1 g   k1 F
m1
a1  8.3m / s 2 (down)
a2 y 

 m2 g   k 2 FN
m1
 m2 g   k 2 F
m2
a2  7.23m / s 2 (down)
c) Why are the accelerations different? The coefficient and force of kinetic friction acting on each
of the blocks is different and this causes the acceleration of each block to be different
15. A block is set on an adjustable ramp and the angle of incline is increased slowly until the block is
observed to start sliding. (a) Solve for the coefficient of static friction, μs, in terms of the maximum
angle of incline, θ, at which the block can remain at rest on the ramp.
Fx  max  0
FN
fs
Fgx  f s  0
Fg sin q   s FN  0
mg sin q   s Fgy  0
Fg2x
mg sin q   s mg cos q  0
mg sin q sin q

mg cos q cos q
 s  tan q
s 
q
Fgy
q
Fg=mg
(b) Considering the fact that μs is almost always less than or equal to one, at what amount of incline
would there be too little friction to prevent virtually anything from sliding?
At q = 90o, since tanq = 1