Homework – Problem Solving Newton’s Laws of Motion 1. What is the weight of a 2.26 kg bag of sugar? W = m g = (2.26 kg)(-9.8 m/s2) = -22.15 N 2. What net force is required to accelerate a 1500 kg car at 3 m/s2? Fnet = m a = (1500 kg)(3 m/s2) = 4500 N = 4.5 kN 3. What net force is required to accelerate a 7500 kg F-14 Tomcat fighter jet off the deck of an aircraft carrier at 9 m/s2? Fnet = m a = (7500 kg)(9 m/s2) = 67500 N = 67.5 kN 4. A 6 kg baby is dropped by a stork 50 meters above a house. Being a quick-thinking baby (obviously a future physicists), the baby takes off her diaper and uses it as a parachute. If the parachute offers an upward force of 35 N on the baby, calculate the net force acting on the baby. What is the acceleration of the baby as she falls with her diaper-parachute open? W = m g = (6 kg)(-9.8 m/s2) = -58.8 N Fnet =  Fy = W + Fdrag = -58.8 N + 35 N = -23.8 N a 5. Fnet  23.8 N   -3.97 m/s2 m 6kg Two people are pushing a stalled car. The mass of the car is 1850 kg. One person applies a force of 275 N to the car, while the other person applies a force of 395 N. Both people are pushing in the same direction. If there are 560 N of friction opposing the car’s motion, what is the acceleration of the car? W = m g = (1850 kg)(-9.8 m/s2) = -18130 N FN = -W = 18130 N Fnet =  Fx = Ff + F1 + F2 = -560 N + 275 N + 395 N = 110 N F 110 N a  net   0.0595 m/s2 = 5.95  10-2 m/s2 m 1850kg Homework – Problem Solving Newton’s Laws of Motion (cont.) 6. A 12 kg sign is pulled upward by a rope with an acceleration of 3 m/s2. Calculate the tension in the rope. W = m g = (12 kg)(-9.8 m/s2) = -117.6 N Fnet = Fy = W + T Fnet = m a = (12 kg)(3 m/s2) = (-117.6 N + T) T = 36 N + 117.6 N = 153.6 N 7. Calculate the acceleration of a 10 kg block if it is placed on a 30 incline plane. Calculate the acceleration if the angle is increased to 60. Ignore friction in both cases. How is the acceleration affected if the mass increases to 100 kg? a = g sin = (-9.8 m/s2)(sin 30) = -4.9 m/s2 a = g sin = (-9.8 m/s2)(sin 60) = -8.5 m/s2 Acceleration is not affected by weight or mass. 8. Calculate the acceleration of a 10 kg block if it is placed on a 30 incline plane. Calculate the acceleration if the angle is increased to 60. The coefficient of friction between the incline and block is 0.25. How is the acceleration affected if the mass increases to 100 kg? a = g(sin –  cos) = (-9.8 m/s2)[sin 30 – (0.25)(cos 30)] = -2.8 m/s2 a = g(sin –  cos) = (-9.8 m/s2)[sin 60 – (0.25)(cos 60)] = -7.3 m/s2 Acceleration is not affected by weight or mass. 9. A 100 kg trunk is lifted straight up with a constant force of 1080 N. Calculate the net force acting on the trunk and its acceleration. W = m g = (100 kg)(-9.8 m/s2) = -980 N Fnet =  Fy = W + FA = -980 N + 1080 N = 100 N F 100 N a  net   1 m/s2 m 100kg Homework – Problem Solving Newton’s Laws of Motion (cont.) 10. A 157 kg barrel needs to be moved across a warehouse. A force of 395 N is applied to the barrel to slide it across the dirt floor. The coefficient of friction between the dirt and barrel is 0.79. Calculate the net force acting on the barrel and its acceleration. W = m g = (157 kg)(-9.8 m/s2) = -1538.6 N FN = -W = 1538.6 N Ff =  FN = (0.79)(1538.6 N) = 1215.5 N Fnet =  Fx = Ff + FA = -1215.5 N + 395 N = -820.5 N F  820.5 N a  net   -5.2 m/s2 m 157kg The applied force is not large enough to move the barrel. Since the net force is negative, the acceleration is zero.