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Example Problems for Newton’s Second Law Answers
1. What force must act on a 50.0-kg mass to give it an
acceleration of 0.30 m/s2?
F = ma
= ( 50.0 kg )( 0.30 m/s2 )
= 15 kg m/s2
= 15 newtons
= 15 N
2. A 1500-kg car starting from rest attains a speed of
25.0 m/s in 50.0 s.
(a) What is the acceleration?
vi = 0
a =
vf = 25.0 m/s
vf - vi
t
=
t = 50.0 s
25.0 m/s - 0
50.0 s
a = 0.50 m/s2
(b) What force is acting on it?
F = m a = ( 1500 kg )( 0.50 m/s2 )
F = 750 N
a = ?
3. A rocket has a mass of 5.00 x 104 kg.
(a) What does the rocket weigh?
Wt. = m g = ( 5.00 x 104 kg )( 9.80 m/s2 )
Wt. = 4.90 x 105 N
Wt. = 490 000 N
(b) What is the minimum force that must
be supplied by the engines in order to
lift the rocket off the pad?
FA
Weight acts downward
Applied force by engines acts upward
to counteract gravity
Wt. = - 490 000 N
(b) What is the minimum force that must
be supplied by the engines in order to
lift the rocket off the pad?
FA
Weight acts downward
Applied force by engines acts upward
to counteract gravity
Remove launch pad:
Wt. = - 490 000 N
(b) What is the minimum force that must
be supplied by the engines in order to
lift the rocket off the pad?
FA
Weight acts downward
Applied force by engines acts upward
to counteract gravity
Remove launch pad:
Wt. = - 490 000 N
If applied force is less than weight,
gravity wins and rocket falls
(b) What is the minimum force that must
be supplied by the engines in order to
lift the rocket off the pad?
FA
Weight acts downward
Applied force by engines acts upward
to counteract gravity
Remove launch pad:
Wt. = - 490 000 N
If applied force is greater than
weight, FA wins and rocket goes up
(b) What is the minimum force that must
be supplied by the engines in order to
lift the rocket off the pad?
FA
Weight acts downward
Applied force by engines acts upward
to counteract gravity
Remove launch pad:
Wt. = - 490 000 N
If FA = Wt., rocket will “hover”
Minimum force to lift rocket off the pad is equal
to the weight
FA = + 4.90 x 105 N
(c) What additional force must be supplied by the engines in
order to accelerate the rocket upward at 15.0 m/s2?
Additional force will be a net
force that accelerates the rocket
FA
Fnet
a = +15.0
Fnet = m a
m/s2
= ( 50 000 kg )( + 15.0 m/s2 )
Fnet = + 750 000 N
Wt. = - 490 000 N
m = 50 000 kg
(d) What is the total force exerted by the engines?
FA = ?
Total force does two things:
(1) counteracts the weight and
(2) accelerates the rocket
Fnet = +750 000 N
Total applied force will be the
a = +15.0 m/s2 sum of their magnitudes:
FA =
Wt. = - 490 000 N
750 000 N
+ 490 000 N
FA = + 1 240 000 N
= + 1.24 x 106 N
(d) What is the total force exerted by the engines?
A more general (and mathematical) approach:
A net force is always the vector sum
of all the forces acting in a
Fnet = +750 000 N
particular dimension:
Fnet = F1 + F2 + …
FA = ?
In this case, the dimension is the
vertical dimension and the individual
forces are the applied force and the
weight
Fnet = FA + Wt.
Wt. = - 490 000 N
(d) What is the total force exerted by the engines?
Fnet = FA + Wt.
+ 750 000 N = FA + ( - 490 000 N )
Fnet = +750 000 N
+ 490 000 N
+ 490 000 N
FA = + 1 240 000 N
FA = ?
= + 1.24 x 106 N
Wt. = - 490 000 N
4. A rocket having a mass of 100 kg is pushed upward by its
engine with a force of 1470 N.
(a) What is the net upward force
acting upon it?
FA = +1470 N
Fnet = ?
Total force of engines must
always counteract the weight
Wt. = m g = ( 100 kg )( 9.8 m/s2 )
Wt. = 980 N
m = 100 kg
4. A rocket having a mass of 100 kg is pushed upward by its
engine with a force of 1470 N.
(a) What is the net upward force
acting upon it?
FA = +1470 N
Fnet = ?
Since the force from the engines is
greater than the weight, there is a
net upward force
The net force is the difference
between FA and the weight
Fnet = ( 1470 N ) - ( 980 N )
Fnet = 490 N
m = 100 kg
Wt. = 980 N
4. A rocket having a mass of 100 kg is pushed upward by its
engine with a force of 1470 N.
(a) What is the net upward force
acting upon it?
FA = +1470 N
Fnet = ?
Fnet = FA + Wt.
= ( 1470 N ) + ( - 980 N )
Fnet = 490 N
m = 100 kg
Wt. = - 980 N
4. A rocket having a mass of 100 kg is pushed upward by its
engine with a force of 1470 N.
(b) What acceleration does this
force produce?
Fnet = 490 N
Fnet = m a
m
m
a =
=
Fnet
=
m
490 N
100 kg
490 kg m/s2
100 kg
a = 4.9 m/s2
FA = +1470 N
m = 100 kg
Wt. = - 980 N
5. A box of mass 25 kg is pulled across a floor. It requires a
force of 120 N to move the box at constant velocity.
Calculate the coefficient of friction μ.
μ =
Ff
FN
Free-body diagram
(diagram showing all
the forces on an
object)
FA = 120 N
Ff
Wt.
FN
For object on horizontal surface, FN equals the Wt.
FN = Wt. = m g = ( 25 kg )( 9.8 m/s2 )
FN = 245 N
5. A box of mass 25 kg is pulled across a floor. It requires a
force of 120 N to move the box at constant velocity.
Calculate the coefficient of friction μ.
μ =
Ff
FN
m = 25 kg
Ff
μ = ?
Wt.
constant velocity
a = 0
Ff = FA
Ff = 120 N
FA = 120 N
245 N
Fnet = 0
5. A box of mass 25 kg is pulled across a floor. It requires a
force of 120 N to move the box at constant velocity.
Calculate the coefficient of friction μ.
μ =
=
Ff
FN
120 N
245 N
μ = 0.49
m = 25 kg
Ff = 120 N
μ = ?
Wt.
245 N
FA = 120 N
6. If a box has a mass of 50 kg and the coefficient of friction
is 0.24, find the force of friction.
FN μ =
Ff
FN
FN
m = 50 kg
Ff
μ = 0.24
Wt.
F f = μ FN
= ( 0.24 )( 490 N )
Ff = 117.6 N
Ff = 120 N
FA
FN
FN = Wt. = m g
= ( 50 kg )( 9.8 m/s2 )
= 490 N
7. The coefficient of starting friction st = 0.65 for steel on a
particular wood floor. What force is needed to begin a steel
box that weighs 2500 N moving across a floor?
Force to begin box
moving must
overcome friction
Ff = μ FN
FA = ?
Ff
μ = 0.65
Wt.
FN
7. The coefficient of starting friction st = 0.65 for steel on a
particular wood floor. What force is needed to begin a steel
box that weighs 2500 N moving across a floor?
Force to begin box
moving must
overcome friction
Ff = μ FN
= ( 0.65 )( 2500 N )
Ff = 1625 N
Ff = 1600 N
FA = ?
Ff
μ = 0.65
Wt.
FN
FN = Wt. = 2500 N
8. The coefficient of sliding friction sl = 0.55 for the steel
box in the previous problem. What force is necessary to
keep the box moving at constant speed across the floor?
FA = ?
Ff
μ = 0.65
Wt.
FN
FN = Wt. = 2500 N
8. The coefficient of sliding friction sl = 0.55 for the steel
box in the previous problem. What force is necessary to
keep the box moving at constant speed across the floor?
FA = ?
Ff
μ = 0.65
Wt.
FN
FN = Wt. = 2500 N
constant speed
a = 0
FA = Ff
Fnet = 0
8. The coefficient of sliding friction sl = 0.55 for the steel
box in the previous problem. What force is necessary to
keep the box moving at constant speed across the floor?
FA = ?
Ff
Ff = μ FN
= ( 0.55 )( 2500 N )
Ff = 1375 N
Ff = 1400 N
μ = 0.65
Wt.
FN
FN = Wt. = 2500 N
FA = Ff
9. Calculate the acceleration of the box in the diagram below.
FA = + 30 N
Ff = - 20 N
4.0 kg
Fnet
Fnet = FA + Ff = ( + 30 N ) + ( - 20 N )
Fnet = + 10 N
Fnet = m a
m
m
a =
Fnet
m
=
+ 10 N
4.0 kg
= a = + 2.5 m/s2
10. An object of mass 500 kg is resting on the floor. The
coefficient of friction μ = 0.40. Calculate the force
necessary to accelerate the object at 2.4 m/s2.
Ff
FA = ?
500 kg
Fnet
a = 2.4 m/s2
μ = 0.40
Wt.
FN
Ff = μ FN = μ Wt. = μ m g
= ( 0.40 )( 500 kg )( 9.8 m/s2 )
Ff = 1960 N
10. An object of mass 500 kg is resting on the floor. The
coefficient of friction μ = 0.40. Calculate the force
necessary to accelerate the object at 2.4 m/s2.
Ff = 1960 N
FA = ?
500 kg
Fnet
a = 2.4 m/s2
μ = 0.40
Wt.
FN
Fnet = m a = ( 500 kg )( 2.4 m/s2 ) = 1200 N
Applied force: (1) overcomes friction
and (2) accelerates the box
FA = ( 1960 N ) + ( 1200 N ) = FA = 3160 N
FA = 3200 N
10. An object of mass 500 kg is resting on the floor. The
coefficient of friction μ = 0.40. Calculate the force
necessary to accelerate the object at 2.4 m/s2.
Ff = - 1960 N
FA = ?
500 kg
Fnet = + 1200 N
a = 2.4 m/s2
μ = 0.40
Wt.
FN
Fnet = FA + Ff
+ 1200 N = FA + ( - 1960 N )
+ 1960 N
+ 1960 N
+ 3160 N = FA
FA = + 3200 N
11. Two boxes of mass 20 kg and 15 kg are connected by a rope. The
rope is suspended over a frictionless pulley on the edge of a table.
The coefficient of friction between the 20-kg box and the table
is 0.25. Find the acceleration of the boxes.
The force that accelerates the
boxes comes from the
weight of m2.
Wt. = m2 g
= ( 15 kg )( 9.8 m/s2 ) = 147 N
m1
20 kg
μ = 0.25
m2
15 kg
m2 g
11. Two boxes of mass 20 kg and 15 kg are connected by a rope. The
rope is suspended over a frictionless pulley on the edge of a table.
The coefficient of friction between the 20-kg box and the table
is 0.25. Find the acceleration of the boxes.
The force is transmitted to m1
by the rope
Ff = μ FN
Ff
m1
20 kg
FA = 147 N
μ = 0.25
m2
15 kg
FN is the weight of the 20-kg box
= m1 g = ( 20 kg )( 9.8 m/s2 ) = 196 N
Ff = μ FN = ( 0.25 )( 196 N ) = 49 N
147 N
11. Two boxes of mass 20 kg and 15 kg are connected by a rope. The
rope is suspended over a frictionless pulley on the edge of a table.
The coefficient of friction between the 20-kg box and the table
is 0.25. Find the acceleration of the boxes.
Fnet
m1
Ff = - 49 N 20 kg
Fnet = FA + Ff
= ( + 147 N ) + ( - 49 N )
Fnet = + 98 N
Fnet = m a
m
m
Fnet
a =
m
=
FA = + 147 N
μ = 0.25
m2
15 kg
Both masses accelerate,
so total mass must be
used in equation
+ 98 N
( 20 + 15 ) kg
= a = + 2.8 m/s2
147 N