Name
Date
Pd
Net Force Particle Model Worksheet 4:
Newton's 2nd Law and Component Forces
1. A rollercoaster car, 300 kg with passengers, accelerates down a 65° hill. We will assume that
friction is small enough that it can be ignored.
a. Draw a force diagram for the system of
car and riders.
Fnet
F
N
F
┴
6
g 5
F┴ °
g║
║
║
║
F
g
b. Determine the value of the component of the gravitational force parallel to the hill.
N
Fgpar  (sin )Fg  (sin65 )(300kg)(10 kg
)  2720N
c. What is the acceleration of the rollercoaster down the hill?
a
Fnet Fgpar 2720N


 9.1 sm2
m
m
300kg
d. If they are traveling 4 m/s at the time shown in the picture, how fast will they be traveling
at the end of the straight stretch, 20 m later?
v f2  v i2  2ax  v f2  2ax  v i2
v f2  2(9.1 sm2 )(20m)  (4 ms )2  v f2  364 ms 2  16 ms 2
2
v f2  380 ms 2  v   380 ms 2  20 ms
2
2
2
2. A worker pushes a 7 kg shipping box along a roller track. Assume friction is small enough to
be ignored because of the rollers. The worker's push is 25 N directed down and to the right at
an angle of 20°.
a. Draw a force diagram for the block.y
F
N
2
0
°
25
N
F
g
b. Determine the horizontal-component of the worker's push.
Fx  cos20 (25N)  23.5N
c. Write a net force equation for the horizontal forces on the block.
Fnet = Fx = 23.5N
d. Determine the acceleration of the block.
a
Fnet 23.5N

 3.4 sm2
m
7.0kg
e. Determine the normal force on the block.
FN  Fy  Fg  0
N
FN  sin20 (25N)  (7kg)10 kg
0
FN  8.6N  70N  79N
x
3. A 70 kg box is pulled by a 400 N force at an angle of 30° to the horizontal. The force of
kinetic friction is 75 N.
a. Draw a quantitative force diagram
for the box.
N
Fg  mg  70kg(10 kg
)  700N
Fx  cos30 (400N)  346N
Fy  sin30 (400N)  200N
FN  Fy  Fg  0  FN  200N  700N  0
y
FN =
FN  500N
500N
400
Ny=
F
Ffk=
x
200NFx=
75N
346N
b. Determine the acceleration of the box.
Fg =
700N
F
F  Ffk 346N  75N 271N
a  net  x


 3.9 sm2
m
m
70kg
70kg
or 3.9
m
s2
to the left
4. Below is a picture of an Atwood's Machine: two masses attached to a frictionless, massless
pulley (pretty neat how physicists dream up equipment like this, huh?). The mass of block A
is 5.0 kg, and the mass of B is 2.0 kg.
a. Determine the acceleration of the system when the blocks are
released.
For the system it is convenient to make the coordinate
system follow the string. So positive is up for B and down
for A. Therefore A is pulled by the earth in the positive
direction and B in the negative. The string pulls in opposite
directions equally so it exerts no net force on the system
N
N
)  2.0kg(10 kg
) 30N
Fnet Fg onA  Fg onB 5.0kg(10 kg
a



 4.3 sm2
m
total mass
5kg  2kg
7.0kg
b. How long will it take for block A to fall 2.0 m?
y  v i t  21 at 2  y  0  21 at 2  t 
2y
t 
a
2(2.0m)
 0.96s
4.3 sm2
5. Two lovers are parked 10.0 m from the edge of a cliff in a car whose mass, including that of
the occupants is 1000 kg. A jealous suitor ties a rope to the car's bumper and a 50 kg rock to
the other end of the rope. He then lowers the rock over the edge of the cliff, and the car,
which is in neutral, accelerates toward the edge. (Note the similarity to the modified
Atwood's machine lab, and ignore frictional effects.)
a. Draw force diagrams for the rock
and the car:
F
N
F FT
T
Fg
Fg
Illustration By Blair Turner, class of 2003.
b. Determine the acceleration of the car towards the edge.
The net force on the system is the force of the earth's gravity on the rock. The two
tensions cancel out and the force of gravity on the earth is canceled out by the normal
force of the earth on the car.
 
N
50kg 10 kg
Fnet weight of rock
500N
a



 0.48 sm2
m
total mass
1000kg  50kg 1050kg
c. How long do the lovers have to apply the brakes before they go over the edge?

x  v i t  21 at 2  x  0  21 at 2  t 
2x
t 
a
2(10m)
 6.5s
.48 sm2