Download 1. A race car has a mass of 710 kg. It starts from rest and travels 40.0

1. A race car has a mass of 710 kg. It starts from rest and travels 40.0m in 3.0s. The car is uniformly accelerated during
the entire time. How big is the net force acting on the car? Make a quantitative force diagram. Write a net force
equation for the axis along which forces are not balanced.
vi = 0.0 m/s vf = ___ x = 40 m
t = 3.0 s
a = ___
Fnet = ____
data
time velocity position
0.0 0.0
0.0
1.0 ____
___
2.0 ____ ____
3.0 26.6 _40_
FN
x 40  0m
v 

 13 .3 m s
t
3s
v v
v  i f  v f  2v v i
2
v f  2 13 .3 m s  26 .6 m s

Ff,car,road
m
v 26 .6  0 ms

 8.9 s
t
3s
s
F
a  net  Fnet  ma
m


Froad  710 k g 8.9 m 2  6320 N

s 
a

Fg

2. Suppose that a 1000 kg car istraveling at 25 m/s (55 mph). Its brakes can apply a force of 5000 N. What is the
minimum distance required for the car to stop? Make a quantitative force diagram. Write a net force equation for
the axis along which forces are not balanced.
vi = 55 m/s
vf = 0.0 m/s
a = ___
Fbrakes = 5000 N
Fbrakes
a
Fnet
m
a
a
5000 N
1000 kg
a  5.0
m
s2
Ff
m
FN
Fbrake
or  5.0
m
s
each second
3. A 65 kg person dives into the water from the 10 m platform.
a.
What is her speed as she enters the water?
vi = 0.0 m/s vf = ____
y = -10 m
v f2  v i2  2ay
v f2  0  2( 10 sm2 )( 10m)
v f2  200 ms 2  v f   200  14 ms
2
a = -10 m/s/s
Fg
b.
She comes to a stop 4.0 m below the surface of the water. Find the force on the swimmer by the water.
vi = 14.14 m/s
vf = 0.0 m/s y = - 4.0 m
a = ____
v f2  v i2  2ay
a
a
a
v f2 v i2
2 y
0 ( 14.14 ms )2
2( 4 m )

2
0(200 m2
s
)
8 m
m
a  25 sm2  25 s s

Fnet
m
Fwater Fg
m
 Fwater  Fg  ma
Fwater
Fw  ma  Fg
N
Fwater  65kg (25 sm2 )  65kg (10 kg
)  1625N  650N  2275N  2300N
Fnet
Fg
4. During a head-on collision, a passenger in the front seat of a car accelerates from 13.3 m/s (30 miles/hour) to rest in
0.10 s.
a.
Calculate the acceleration of the passenger.
vi = 13.3 m/s vf = 0.0 m/s
t = 0.10 s
a = ____
assume passenger in left car and to right is positive
a
v
t
a
m
0  13.3 ms
 133 sm2  133 ss
.10s
b.
The driver of the car holds out his arm to keep his 25 kg child (who is not wearing a seat belt) from smashing into
the dashboard. How much force must he exert on the child?
vi = 13.3 m/s
a
Fnet
m

FN
m
vf = 0.0 m/s
t = 0.10 s
 FN  ma
a = - 133 m/s/s (from part a.)
m = 25 kg
Fnet
FN  25kg ( 133 sm2 )  3325N
Farm
c.
What is the weight of the child?
N
Fg  mg  25kg (10 kg
)  250N
d.
Convert the forces in parts b and c from Newtons to pounds. (1 lb = 4.45N). What are the chances the driver will
be able to stop the child?
1lb
)  747 pounds
4.45N
1lb
250N(
)  56 pounds
4.45N
3325N(
The driver has no chance of stopping the child.
6.
The following questions refer to the motion of a baseball.
a.
While being thrown, a net force of 132 N acts on a baseball (mass = 140 g) for a period of 4.5 x 10-2 sec.
What is the magnitude of the change in momentum of the ball?
F t  mv  F t  132N (0.045s )  5.9N s  5.9 kgs m
b.
If the initial speed of the baseball is v = 0.0 m/s, what will its speed be when it leaves the pitcher's hand?
mv  5.9 kgs m  v 
5.9 kgs m
5.9 kgs m

 42 ms
1kg
140g 1000
0.140
kg
g
c.
When the batter hits the ball, a net force of 1150 N, opposite to the direction of the ball's initial motion, acts
on the ball for 9.0 x 10-3 s during the hit. What is the final velocity of the ball?
F t  mv  F t  1150N(0.0090s )  (0.140kg )v  v  74 ms
v  v f  v i  v f  v  v i  v f  74 ms  42 ms  32 ms
d.
vi
How large is the force the ball exerts on the bat? Explain.
The ball exerts a force on the bat of 1150N. Ball on bat and bat on ball are a 3rd Law pair.
7.
A rocket, weighing 4.36 x 104N, has an engine that provides an upward force of 1.2 x 105N. It reaches a
maximum speed of 860 m/s.
Fengine
a.
Draw a force diagram for the rocket.
Fg
b. For how much time must the engine burn during the launch in order to reach this
speed?
4.3510 N
Fg
(860 ms )
N
10 kg
mv
g v
Fnet t  mv  t 


 49s
Fnet
1.2  105 N  4.35  104 N
7.65  104 N
4