1. AP Forces Review Problem Set Answers
1. m = kg F = N a = ?
F = m a F
25.0 N
a = = m
0.012 kg
a = m/s2

2. 2. m = kg FA = 3.5 N μ = ? Ff
If block is pulled at constant velocity,
then FA = Ff = 3.5 N
μ = FN
On horizontal surface (table),
FN = Wt. = mg
= ( kg )( 9.8 m/s2 )
= N Ff
3.5 N
μ = =
= μ = 0.65 FN
5.39 N

3. 3. m = 300 kg μ = 0.42
(a) force required to move block at constant velocity = ?
If block is pulled at constant velocity, then FA = Ff
Ff = μ FN
= μ Wt.
= μ m g
= ( 0.42 )( 300 kg )( 9.8 m/s2 )
Ff = FA = N

4. 3. m = 300 kg μ = Ff = N
(b) total force required to accelerate at 1.6 m/s2 = ?
To accelerate cart:
Fnet = m a
= ( 300 kg )( 1.6 m/s2 )
Fnet = 480 N
Total force = N N
Total force = N

5. 4. acceleration of 45.0 kg block = ?
Weight of m2 accelerates system
20.0 kg
m2 g = ( 20.0 kg )( 9.8 m/s2 )
= 196 N
m2 g
Force is transmitted to m1 by string
F = m a F
196 N
Force accelerates
both masses
a = = m
( kg )
a = m/s2

6. 5. μ = 0.15 acceleration of 45.0 kg block = ?
Again, weight of m2
accelerates system Ff
45.0 kg
μ = 0.15
Now, however, there is
friction present
20.0 kg
Ff = μ FN
= μ m1 g
m2 g
= ( 0.15 )( 45.0 kg )( 9.8 m/s2 )
Ff = N

7. 5. μ = 0.15 acceleration of 45.0 kg block = ?
FA = 196 N
Ff = N
45.0 kg
Fnet = FA + Ff
μ = 0.15
= ( N ) + ( N )
20.0 kg
Fnet = N
m2 g
Fnet N
Then
a = = m
( kg )
a = 2.0 m/s2

8. Weight of M accelerates system
6. Find M.
FA = 9.8 M Ff
20.0 kg
μ = 0.25
Weight of M accelerates system M
Wt. = M g
= 9.8 M
a = 0.50 m/s2
Friction force:
Ff = μ FN
= μ m1 g
9.8 M
= ( 0.25 )( 20 kg )( 9.8 m/s2 )

9. Weight of M accelerates system
6. Find M.
FA = 9.8 M
Ff = N
20.0 kg
μ = 0.25
Weight of M accelerates system M
Wt. = M g
= 9.8 M
a = 0.50 m/s2
Friction force:
Ff = μ FN
= μ m1 g
9.8 M
= ( 0.25 )( 20 kg )( 9.8 m/s2 )
Ff = 49 N
Fnet = FA + Ff
= ( M ) + ( - 49 )
Fnet
= 9.8 M - 49

10. Also, Fnet accelerates the masses at 0.50 m/s2
6. Find M.
FA = 9.8 M
Ff = N
20.0 kg
Fnet
= 9.8 M
μ = 0.25
Also, Fnet accelerates the masses at 0.50 m/s2 M
(both masses)
a = 0.50 m/s2
Fnet = m a
9.8 M
= ( M + 20 kg )( 0.50 m/s2 )
= M + 10
Set equal to each other:
9.8 M
= M + 10
Solve for M

11. 6. Find M.
FA = 9.8 M
Ff = N
20.0 kg
μ = 0.25 M
9.8 M
= M + 10
a = 0.50 m/s2
9.8 M
= M + 59
9.8 M
9.3 M
= 59
M = 6.3 kg

12. 7. Find the acceleration of m2 .
m1 = 1.5 kg m2 = 2.5 kg
m1 g = ( 1.5 kg )( 9.8 m/s2 ) = N
m2 g = ( 2.5 kg )( 9.8 m/s2 ) = N m1 m2
Fnet
Fnet = ( 24.5 N ) - ( 14.7 N ) = 9.8 N
m1 g
m2 g
Fnet = m a
Fnet
9.8 N
Force accelerates
both masses
a = = m
( kg )
a = m/s2

13. 8. m = 1000 kg Total thrusting force = 25 000 N
FA = N
Part of FA is used to overcome the weight
Fnet = ?
Wt. = m g
= ( 1000 kg )( 9.8 m/s2 )
Wt. = N
Fnet = FA + Wt.
m = 1000 kg
= ( N ) + ( N )
Fnet = N
Wt. = N

14. 8. m = 1000 kg Total thrusting force = 25 000 N
FA = N
Fnet = N
Fnet = N
Then
Fnet = m a
Fnet
a = m
m = 1000 kg N =
1000 kg
Wt. = N
a = m/s2

15. 9. Mass decreases by 10% every minute (because fuel is consumed)
FA = N
What is the mass after 5 minutes?
Mass after 1 minute = 90% of 1000 kg
Mass after 1 minute
= 900 kg
Mass after 2 minutes = 90% of 900 kg
Mass after 2 minutes
= 810 kg
m = 1000 kg
Mass after 3 minutes
= 729 kg
Mass after 4 minutes
= kg
Wt.
Mass after 5 minutes
= kg
Then Wt. = m g
= ( kg )( 9.8 m/s2 )
Wt. = N

16. 9. Mass decreases by 10% every minute a after 5 minutes = ?
Fnet = FA + Wt.
FA = N
= ( N ) + ( N )
Fnet = ?
Fnet = N
Then
Fnet = m a
Fnet
a = m
m = 1000 kg N = kg
Wt. = N
a = m/s2