Example Problems for Newton’s Second Law Answers 1. What force must act on a 75.0-kg mass to give it an acceleration of 1.2 m/s2? F = m a = ( 75.0 kg )( 1.2 m/s2 ) = 90 kg m/s2 = 90 newtons = 90 N

2. A 2500-kg car starting from rest attains a speed of 15.0 m/s in 30.0 s. (a) What is the acceleration? vi = 0 vf = 15.0 m/s t = 30.0 s a = ? vf - vi 15.0 m/s - 0 a = = t 30.0 s a = 0.50 m/s2 (b) What force is acting on it? F = m a = ( 2500 kg )( 0.50 m/s2 ) F = 1250 N

3. A rocket has a mass of 575 kg. (a) What does the rocket weigh? Wt. = m g = ( 575 kg )( 9.80 m/s2 ) Wt. = 5635 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. = - 5635 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. = - 5635 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. = - 5635 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. = - 5635 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. = - 5635 N If FA = Wt., rocket will “hover” Minimum force to lift rocket off the pad is equal to the weight FA = + 5635 N

(c) What additional force must be supplied by the engines in order to accelerate the rocket upward at 12.0 m/s2? Additional force will be a net force that accelerates the rocket FA Fnet Fnet = m a = ( 575 kg )( + 12.0 m/s2 ) a = +12.0 m/s2 Fnet = + 6900 N Wt. = - 5635 N m = 575 kg

(d) What is the total force exerted by the engines? Total force does two things: (1) counteracts the weight and (2) accelerates the rocket FA = ? Fnet = + 6900 N Total applied force will be the sum of their magnitudes: a = + 12.0 m/s2 FA = 5635 N + 6900 N Wt. = - 5635 N FA = + 12 535 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 particular dimension: Fnet = + 6900 N FA = ? Fnet = F1 + F2 + … In this case, the dimension is the vertical dimension and the individual forces are the applied force and the weight Fnet = FA + Wt. Wt. = - 5635 N

(d) What is the total force exerted by the engines? Fnet = FA + Wt. + 6900 N = FA + ( - 5635 N ) Fnet = + 6900 N + 5635 N + 5635 N FA = + 12 535 N FA = ? Wt. = - 5635 N

4. A rocket having a mass of 1800 kg is pushed upward by its engine with a force of 25 000 N. FA = +25 000 N (a) What is the net upward force acting upon it? Fnet = ? Total force of engines must always counteract the weight Wt. = m g = ( 1800 kg )( 9.8 m/s2 ) Wt. = 17 640 N m = 1800 kg

4. A rocket having a mass of 1800 kg is pushed upward by its engine with a force of 25 000 N. FA = +25 000 N (a) What is the net upward force acting upon it? 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 m = 1800 kg Fnet = ( 25 000 N ) - ( 17 640 N ) Wt. = 17 640 N Fnet = 7360 N

4. A rocket having a mass of 100 kg is pushed upward by its engine with a force of 1470 N. FA = +25 000 N (a) What is the net upward force acting upon it? Fnet = ? Fnet = FA + Wt. = ( 25 000 N ) + ( - 17 640 N ) Fnet = 7360 N m = 1800 kg Wt. = - 17 640 N

4. A rocket having a mass of 1800 kg is pushed upward by its engine with a force of 25 000 N. (b) What acceleration does this force produce? FA = +25 000 N Fnet = m a Fnet = 7360 N m m Fnet 7360 N a = = m 1800 kg 7360 kg m/s2 m = 1800 kg = 1800 kg Wt. = - 17 640 N a = 4.09 m/s2

5. A box of mass 6.4 kg is pulled across a floor. It requires a force of 36 N to move the box at constant velocity. Calculate the coefficient of friction μ. Ff μ = FN FA = 36 N Free-body diagram (diagram showing all the forces on an object) Ff FN Wt. For object on horizontal surface, FN equals the Wt. FN = Wt. = m g = ( 6.4 kg )( 9.8 m/s2 ) FN = 62.72 N

5. A box of mass 6.4 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. m = 6.4 kg FA = 36 N Ff Ff μ = FN μ = ? 62.72 N Wt. constant velocity a = 0 Fnet = 0 Ff = FA Ff = 36 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. m = 6.4 kg FA = 36 N Ff μ = Ff = 36 N FN μ = ? 62.72 N 36 N Wt. = 62.72 N μ = 0.57

6. If a box has a mass of 24 kg and the coefficient of friction is 0.40, find the force of friction. m = 24 kg FA Ff FN Ff FN μ = FN μ = 0.40 Wt. FN Ff = μ FN FN = Wt. = m g = ( 0.40 )( 235.2 N ) = ( 24 kg )( 9.8 m/s2 ) Ff = 94.08 N = 235.2 N Ff = 94 N

7. The coefficient of starting friction st = 0.45 for steel on a particular wood floor. What force is needed to begin a steel box that weighs 1600 N moving across a floor? FA = ? Force to begin box moving must overcome friction Ff μ = 0.65 Wt. FN Ff = μ FN

7. The coefficient of starting friction st = 0.45 for steel on a particular wood floor. What force is needed to begin a steel box that weighs 1600 N moving across a floor? FA = ? Force to begin box moving must overcome friction Ff μ = 0.45 Wt. FN Ff = μ FN FN = Wt. = 1600 N = ( 0.45 )( 1600 N ) Ff = 720 N

8. The coefficient of sliding friction sl = 0.36 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 = 0.36 Wt. FN FN = Wt. = 1600 N

8. The coefficient of sliding friction sl = 0.36 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 = 0.36 Wt. FN FN = Wt. = 1600 N constant speed a = 0 Fnet = 0 FA = Ff

8. The coefficient of sliding friction sl = 0.36 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 μ = 0.36 Wt. FN = ( 0.36 )( 1600 N ) FN = Wt. = 1600 N Ff = 576 N FA = Ff

9. Calculate the acceleration of the box in the diagram below. FA = + 120 N Ff = - 80 N 5.0 kg Fnet Fnet = FA + Ff = ( + 120 N ) + ( - 80 N ) Fnet = + 40 N Fnet = m a m m Fnet + 40 N a = = = a = + 8.0 m/s2 m 5.0 kg

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. FA = ? Ff 500 kg Fnet μ = 0.40 a = 2.4 m/s2 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. FA = ? Ff = 1960 N 500 kg Fnet μ = 0.40 a = 2.4 m/s2 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

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. FA = ? Ff = - 1960 N 500 kg Fnet = + 1200 N μ = 0.40 a = 2.4 m/s2 Wt. FN Fnet = FA + Ff + 1200 N = FA + ( - 1960 N ) + 1960 N + 1960 N + 3160 N = FA

11. A 6.0-kg mass and a 14.0-kg mass are attached to a lightweight cord that passes over a frictionless pulley. The masses are left free to move. Find the acceleration of the 6.0-kg mass. F1 = Wt. = m1 g = ( 6.0 kg )( 9.8 m/s2 ) = 58.8 N 6.0 kg 14 kg F2 = Wt. = m2 g = ( 14.0 kg )( 9.8 m/s2 ) F1 = 137.2 N F2

11. A 6.0-kg mass and a 14.0-kg mass are attached to a lightweight cord that passes over a frictionless pulley. The masses are left free to move. Find the acceleration of the 6.0-kg mass. Fnet = 137.2 N - 58.8 N Fnet = 78.4 N Fnet = m a 6.0 kg m m 14 kg Fnet + 78.4 N a = = Fnet 58.8 N m ( 6.0 kg + 14 kg ) a = + 3.92 m/s2 137.2 N

12. Two boxes of mass 4.0 kg and 1.5 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 4.0-kg box and the table is 0.30. Find the acceleration of the boxes. m1 The force that accelerates the boxes comes from the weight of m2. 4 kg μ = 0.30 m2 Wt. = m2 g = ( 1.5 kg )( 9.8 m/s2 ) = 14.7 N 1.5 kg m2 g

12. Two boxes of mass 4.0 kg and 1.5 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 4.0-kg box and the table is 0.30. Find the acceleration of the boxes. m1 The force is transmitted to m1 by the rope FA = 14.7 N Ff 4 kg μ = 0.30 Ff = μ FN m2 1.5 kg FN is the weight of the 4-kg box = m1 g = ( 4.0 kg )( 9.8 m/s2 ) = 39.2 N 14.7 N Ff = μ FN = ( 0.30 )( 14.7 N ) = 11.76 N

12. Two boxes of mass 4.0 kg and 1.5 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 4.0-kg box and the table is 0.30. Find the acceleration of the boxes. Fnet m1 FA = + 14.7 N Ff = - 11.76 N Fnet = FA + Ff 4 kg = ( + 14.7 N ) + ( - 11.76 N ) μ = 0.30 m2 1.5 kg Fnet = + 2.94 N Both masses accelerate, so total mass must be used in equation Fnet = m a 14.7 N m m Fnet + 2.94 N a = = = a = + 0.53 m/s2 m ( 4.0 + 1.5 ) kg