AP Forces Review Problem Set Answers 1. m = 0.012 kg F = 25.0 N a = ? F = m a F 25.0 N a = = m 0.012 kg a = 2083 m/s2

2. m = 0.550 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 = ( 0.550 kg )( 9.8 m/s2 ) = 5.39 N Ff 3.5 N μ = = = μ = 0.65 FN 5.39 N

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 = 1235 N

3. m = 300 kg μ = 0.42 Ff = 1235 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 = 1235 N + 480 N Total force = 1715 N

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 ( 45 + 20 kg ) a = 3.02 m/s2

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 = 66.15 N

5. μ = 0.15 acceleration of 45.0 kg block = ? FA = 196 N Ff = - 66.15 N 45.0 kg Fnet = FA + Ff μ = 0.15 = ( + 196 N ) + ( - 66.15 N ) 20.0 kg Fnet = 129.85 N m2 g Fnet 129.85 N Then a = = m ( 45 + 20 kg ) a = 2.0 m/s2

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 )

Weight of M accelerates system 6. Find M. FA = 9.8 M Ff = - 49 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 = ( + 9.8 M ) + ( - 49 ) Fnet = 9.8 M - 49

Also, Fnet accelerates the masses at 0.50 m/s2 6. Find M. FA = 9.8 M Ff = - 49 N 20.0 kg Fnet = 9.8 M - 49 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 ) = 0.50 M + 10 Set equal to each other: 9.8 M - 49 = 0.50 M + 10 Solve for M

6. Find M. FA = 9.8 M Ff = - 49 N 20.0 kg μ = 0.25 M 9.8 M - 49 = 0.50 M + 10 a = 0.50 m/s2 9.8 M = 0.50 M + 59 9.8 M 9.3 M = 59 M = 6.3 kg

7. Find the acceleration of m2 . m1 = 1.5 kg m2 = 2.5 kg m1 g = ( 1.5 kg )( 9.8 m/s2 ) = 14.7 N m2 g = ( 2.5 kg )( 9.8 m/s2 ) = 24.5 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 ( 1.5 + 2.5 kg ) a = 2.45 m/s2

8. m = 1000 kg Total thrusting force = 25 000 N FA = +25 000 N Part of FA is used to overcome the weight Fnet = ? Wt. = m g = ( 1000 kg )( 9.8 m/s2 ) Wt. = 9800 N Fnet = FA + Wt. m = 1000 kg = ( 25 000 N ) + ( - 9800 N ) Fnet = 15 200 N Wt. = - 9800 N

8. m = 1000 kg Total thrusting force = 25 000 N FA = +25 000 N Fnet = 15 200 N Fnet = 15 200 N Then Fnet = m a Fnet a = m m = 1000 kg 15 200 N = 1000 kg Wt. = - 9800 N a = 15.2 m/s2

9. Mass decreases by 10% every minute (because fuel is consumed) FA = +25 000 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 = 656.1 kg Wt. Mass after 5 minutes = 590.45 kg Then Wt. = m g = ( 590.45 kg )( 9.8 m/s2 ) Wt. = 5787 N

9. Mass decreases by 10% every minute a after 5 minutes = ? Fnet = FA + Wt. FA = +25 000 N = ( 25 000 N ) + ( - 5787 N ) Fnet = ? Fnet = 19 213 N Then Fnet = m a Fnet a = m m = 1000 kg 19 213 N = 590.45 kg Wt. = - 5787 N a = 32.5 m/s2