Unit 2: Dynamics Practice Problems

Problem 1: Newton's Second Law and Friction

1. A 10 kg block is pushed across a horizontal surface with a force of 50 N. The coefficient of kinetic friction between the block and the surface is 0.3.
   a) Calculate the frictional force acting on the block.
   b) Determine the net force acting on the block.
   c) Find the acceleration of the block.

Problem 2: Inclined Plane and Tension

1. A 15 kg box is placed on a 30° incline and connected by a light string over a frictionless pulley to a hanging 10 kg mass. Assume no friction on the incline.
   a) Draw a free-body diagram for each mass.
   b) Calculate the acceleration of the system.
   c) Determine the tension in the string.

Problem 3: Circular Motion and Centripetal Force

1. A 2 kg object moves in a circular path of radius 1.5 m at a constant speed of 4 m/s.
   a) Calculate the centripetal force acting on the object.
   b) If the speed of the object doubles, determine the new centripetal force.
   c) Describe how the force keeping the object in circular motion would change if the radius of the path were tripled while keeping the speed constant.

Problem 4: Force Components and Equilibrium

1. A 12 kg sign is suspended by two cables, each making a 40° angle with the horizontal.
   a) Draw a free-body diagram for the sign, showing the tension forces in the cables and the gravitational force.
   b) Calculate the tension in each cable required to keep the sign in static equilibrium.
   c) Explain what would happen to the tension in each cable if the angle with the horizontal were decreased to 30°.

Problem 5: Forces and Acceleration on a System of Masses

1. Two blocks, A and B, with masses of 8 kg and 12 kg, respectively, are connected by a light string on a horizontal, frictionless surface. A force of 60 N is applied horizontally to block A.
   a) Calculate the acceleration of the system.
   b) Determine the tension in the string between the blocks.
   c) If a frictional force of 15 N is introduced on block B, recalculate the acceleration of the system and the tension in the string.

Problem 6: Gravitational Force and Weight

1. An object weighs 800 N on the surface of Earth.
   a) Calculate the object's mass.
   b) Determine the gravitational force on this object at a distance of twice Earth's radius from the Earth's center.
   c) Discuss the relationship between gravitational force and distance from Earth's center in this context.

Problem 7: Drag Force and Terminal Velocity

1. A skydiver of mass 70 kg reaches terminal velocity while falling through the air.
   a) Explain why terminal velocity occurs, considering the forces involved.
   b) If the skydiver’s terminal velocity is 60 m/s, calculate the magnitude of the drag force acting on the skydiver at terminal velocity.
   c) Describe how increasing the skydiver's cross-sectional area would affect the terminal velocity.

Problem 8: Atwood Machine with Friction

1. In an Atwood machine, a 6 kg mass hangs over a pulley and is connected by a light string to a 4 kg mass on a horizontal surface with a coefficient of kinetic friction of 0.2.
   a) Draw a free-body diagram for each mass.
   b) Calculate the acceleration of the system.
   c) Determine the tension in the string between the two masses.

Problem 9: Dynamics of a Pulley System with Multiple Forces

1. A 5 kg block is hanging vertically and connected to a 7 kg block on an incline of 45° by a light string over a frictionless pulley. The incline is frictionless.
   a) Draw a free-body diagram for each block.
   b) Calculate the acceleration of the system.
   c) Find the tension in the string.
   d) Explain what would happen to the acceleration if a small frictional force were introduced on the incline.

Problem 10: Frictional Forces in Different Conditions

1. A 20 kg sled is pulled along a snowy surface with a coefficient of kinetic friction of 0.1 by a rope that makes a 25° angle with the horizontal. The tension in the rope is 75 N.
   a) Calculate the horizontal component of the tension force.
   b) Determine the frictional force acting on the sled.
   c) Calculate the net force on the sled and its resulting acceleration.
   d) Describe how the frictional force would change if the sled moved onto a rougher surface with a higher coefficient of friction.Unit 2: Dynamics Practice Problems

   Problem 1: Newton's Second Law and Friction

   1. A 10 kg block is pushed across a horizontal surface with a force of 50 N. The coefficient of kinetic friction between the block and the surface is 0.3.
      a) Calculate the frictional force acting on the block.
      b) Determine the net force acting on the block.
      c) Find the acceleration of the block.

   Problem 2: Inclined Plane and Tension

   1. A 15 kg box is placed on a 30° incline and connected by a light string over a frictionless pulley to a hanging 10 kg mass. Assume no friction on the incline.
      a) Draw a free-body diagram for each mass.
      b) Calculate the acceleration of the system.
      c) Determine the tension in the string.

   Problem 3: Circular Motion and Centripetal Force

   1. A 2 kg object moves in a circular path of radius 1.5 m at a constant speed of 4 m/s.
      a) Calculate the centripetal force acting on the object.
      b) If the speed of the object doubles, determine the new centripetal force.
      c) Describe how the force keeping the object in circular motion would change if the radius of the path were tripled while keeping the speed constant.

   Problem 4: Force Components and Equilibrium

   1. A 12 kg sign is suspended by two cables, each making a 40° angle with the horizontal.
      a) Draw a free-body diagram for the sign, showing the tension forces in the cables and the gravitational force.
      b) Calculate the tension in each cable required to keep the sign in static equilibrium.
      c) Explain what would happen to the tension in each cable if the angle with the horizontal were decreased to 30°.

   Problem 5: Forces and Acceleration on a System of Masses

   1. Two blocks, A and B, with masses of 8 kg and 12 kg, respectively, are connected by a light string on a horizontal, frictionless surface. A force of 60 N is applied horizontally to block A.
      a) Calculate the acceleration of the system.
      b) Determine the tension in the string between the blocks.
      c) If a frictional force of 15 N is introduced on block B, recalculate the acceleration of the system and the tension in the string.

   Problem 6: Gravitational Force and Weight

   1. An object weighs 800 N on the surface of Earth.
      a) Calculate the object's mass.
      b) Determine the gravitational force on this object at a distance of twice Earth's radius from the Earth's center.
      c) Discuss the relationship between gravitational force and distance from Earth's center in this context.

   Problem 7: Drag Force and Terminal Velocity

   1. A skydiver of mass 70 kg reaches terminal velocity while falling through the air.
      a) Explain why terminal velocity occurs, considering the forces involved.
      b) If the skydiver’s terminal velocity is 60 m/s, calculate the magnitude of the drag force acting on the skydiver at terminal velocity.
      c) Describe how increasing the skydiver's cross-sectional area would affect the terminal velocity.

   Problem 8: Atwood Machine with Friction

   1. In an Atwood machine, a 6 kg mass hangs over a pulley and is connected by a light string to a 4 kg mass on a horizontal surface with a coefficient of kinetic friction of 0.2.
      a) Draw a free-body diagram for each mass.
      b) Calculate the acceleration of the system.
      c) Determine the tension in the string between the two masses.

   Problem 9: Dynamics of a Pulley System with Multiple Forces

   1. A 5 kg block is hanging vertically and connected to a 7 kg block on an incline of 45° by a light string over a frictionless pulley. The incline is frictionless.
      a) Draw a free-body diagram for each block.
      b) Calculate the acceleration of the system.
      c) Find the tension in the string.
      d) Explain what would happen to the acceleration if a small frictional force were introduced on the incline.

   Problem 10: Frictional Forces in Different Conditions

   1. A 20 kg sled is pulled along a snowy surface with a coefficient of kinetic friction of 0.1 by a rope that makes a 25° angle with the horizontal. The tension in the rope is 75 N.
      a) Calculate the horizontal component of the tension force.
      b) Determine the frictional force acting on the sled.
      c) Calculate the net force on the sled and its resulting acceleration.
      d) Describe how the frictional force would change if the sled moved onto a rougher surface with a higher coefficient of friction.

Answer Key for Unit 2: Dynamics Practice ProblemsProblem 1: Newton's Second Law and Friction

1. a) Frictional force:
   fk=μk⋅m⋅g=0.3×10×9.8=29.4 Nf_k = \mu_k \cdot m \cdot g = 0.3 \times 10 \times 9.8 = 29.4 \, \text{N}fk​=μk​⋅m⋅g=0.3×10×9.8=29.4N

   b) Net force:
   Fnet=50 N−29.4 N=20.6 NF_{\text{net}} = 50 \, \text{N} - 29.4 \, \text{N} = 20.6 \, \text{N}Fnet​=50N−29.4N=20.6N

   c) Acceleration:
   a=Fnetm=20.610=2.06 m/s2a = \frac{F_{\text{net}}}{m} = \frac{20.6}{10} = 2.06 \, \text{m/s}^2a=mFnet​​=1020.6​=2.06m/s2

Problem 2: Inclined Plane and Tension

1. a) Free-body diagrams should show gravity, normal force, and tension for each mass.

   b) Acceleration of the system:
   For the 15 kg mass on the incline: m1gsin⁡(30∘)−T=m1am_1 g \sin(30^\circ) - T = m_1 am1​gsin(30∘)−T=m1​a
   For the 10 kg hanging mass: T−m2g=−m2aT - m_2 g = -m_2 aT−m2​g=−m2​a
   Solving these simultaneously gives:
   a=m1gsin⁡(30∘)−m2gm1+m2≈0.98 m/s2a = \frac{m_1 g \sin(30^\circ) - m_2 g}{m_1 + m_2} \approx 0.98 \, \text{m/s}^2a=m1​+m2​m1​gsin(30∘)−m2​g​≈0.98m/s2

   c) Tension in the string:
   T=m2(g+a)=10×(9.8+0.98)≈107.8 NT = m_2(g + a) = 10 \times (9.8 + 0.98) \approx 107.8 \, \text{N}T=m2​(g+a)=10×(9.8+0.98)≈107.8N

Problem 3: Circular Motion and Centripetal Force

1. a) Centripetal force:
   Fc=mv2r=2×421.5=21.3 NF_c = \frac{mv^2}{r} = \frac{2 \times 4^2}{1.5} = 21.3 \, \text{N}Fc​=rmv2​=1.52×42​=21.3N

   b) New centripetal force if speed doubles:
   New speed = 8 m/s
   Fc=2×821.5=85.3 NF_c = \frac{2 \times 8^2}{1.5} = 85.3 \, \text{N}Fc​=1.52×82​=85.3N

   c) Effect of tripling the radius:
   If the radius is tripled, Fc∝1rF_c \propto \frac{1}{r}Fc​∝r1​, so the force would be one-third of the original if speed remains constant.

Problem 4: Force Components and Equilibrium

1. a) The free-body diagram should include gravity and the tension components in both cables at a 40° angle from the horizontal.

   b) Tension in each cable:
   Tcos⁡(40∘)=mg2T \cos(40^\circ) = \frac{mg}{2}Tcos(40∘)=2mg​
   T=12×9.82cos⁡(40∘)≈76.5 NT = \frac{12 \times 9.8}{2 \cos(40^\circ)} \approx 76.5 \, \text{N}T=2cos(40∘)12×9.8​≈76.5N

   c) Effect of decreasing angle to 30°:
   With a smaller angle, cos⁡(30∘)>cos⁡(40∘)\cos(30^\circ) > \cos(40^\circ)cos(30∘)>cos(40∘), so each cable would need to support a greater tension to keep the sign in equilibrium.

Problem 5: Forces and Acceleration on a System of Masses

1. a) Acceleration of the system:
   a=Fmtotal=608+12=3 m/s2a = \frac{F}{m_{\text{total}}} = \frac{60}{8 + 12} = 3 \, \text{m/s}^2a=mtotal​F​=8+1260​=3m/s2

   b) Tension in the string:
   T=mB⋅a=12×3=36 NT = m_B \cdot a = 12 \times 3 = 36 \, \text{N}T=mB​⋅a=12×3=36N

   c) With 15 N friction on block B:
   Net force = 60−15=45 N60 - 15 = 45 \, \text{N}60−15=45N
   New acceleration = 4520=2.25 m/s2\frac{45}{20} = 2.25 \, \text{m/s}^22045​=2.25m/s2
   New tension in the string = T=mB⋅a=12×2.25=27 NT = m_B \cdot a = 12 \times 2.25 = 27 \, \text{N}T=mB​⋅a=12×2.25=27N

Problem 6: Gravitational Force and Weight

1. a) Mass of object:
   m=8009.8≈81.6 kgm = \frac{800}{9.8} \approx 81.6 \, \text{kg}m=9.8800​≈81.6kg

   b) Gravitational force at twice Earth’s radius:
   F=800(2)2=200 NF = \frac{800}{(2)^2} = 200 \, \text{N}F=(2)2800​=200N

   c) Relationship between gravitational force and distance:
   Gravitational force follows the inverse-square law, so it decreases by the square of the distance from the center of Earth.

Problem 7: Drag Force and Terminal Velocity

1. a) Explanation of terminal velocity:
   Terminal velocity occurs when the gravitational force downward is equal to the upward drag force, resulting in no net force and constant velocity.

   b) Drag force at terminal velocity:
   Fdrag=mg=70×9.8=686 NF_{\text{drag}} = mg = 70 \times 9.8 = 686 \, \text{N}Fdrag​=mg=70×9.8=686N

   c) Effect of increasing cross-sectional area:
   Increasing cross-sectional area increases drag force, which would lower the terminal velocity of the skydiver.

Problem 8: Atwood Machine with Friction

1. a) Free-body diagrams should show gravity and friction for each mass.

   b) Acceleration of the system:
   For the hanging mass: m1g−T=m1am_1 g - T = m_1 am1​g−T=m1​a
   For the mass on the surface: T−fk=m2aT - f_k = m_2 aT−fk​=m2​a
   Solving these gives a≈2.8 m/s2a \approx 2.8 \, \text{m/s}^2a≈2.8m/s2

   c) Tension in the string:
   T=m2(a+fk/m2)≈26.4 NT = m_2 (a + f_k/m_2) \approx 26.4 \, \text{N}T=m2​(a+fk​/m2​)≈26.4N

Problem 9: Dynamics of a Pulley System with Multiple Forces

1. a) Free-body diagrams should include gravitational forces, normal force on the incline, and tension.

   b) Acceleration of the system:
   For the hanging block: m1g−T=m1am_1 g - T = m_1 am1​g−T=m1​a
   For the block on the incline: T−m2gsin⁡(45∘)=m2aT - m_2 g \sin(45^\circ) = m_2 aT−m2​gsin(45∘)=m2​a
   Solving yields a≈1.2 m/s2a \approx 1.2 \, \text{m/s}^2a≈1.2m/s2

   c) Tension in the string:
   T=m2gsin⁡(45∘)+m2a≈58.8 NT = m_2 g \sin(45^\circ) + m_2 a \approx 58.8 \, \text{N}T=m2​gsin(45∘)+m2​a≈58.8N

   d) Effect of adding friction on incline:
   A frictional force on the incline would reduce the net force, decreasing the system’s acceleration.

Problem 10: Frictional Forces in Different Conditions

1. a) Horizontal component of tension:
   Tx=Tcos⁡(25∘)=75cos⁡(25∘)≈68 NT_x = T \cos(25^\circ) = 75 \cos(25^\circ) \approx 68 \, \text{N}Tx​=Tcos(25∘)=75cos(25∘)≈68N

   b) Frictional force:
   fk=μk⋅m⋅g=0.1×20×9.8=19.6 Nf_k = \mu_k \cdot m \cdot g = 0.1 \times 20 \times 9.8 = 19.6 \, \text{N}fk​=μk​⋅m⋅g=0.1×20×9.8=19.6N

   c) Net force and acceleration:
   Fnet=Tx−fk=68−19.6=48.4 NF_{\text{net}} = T_x - f_k = 68 - 19.6 = 48.4 \, \text{N}Fnet​=Tx​−fk​=68−19.6=48.4N
   a=Fnetm=48.420≈2.42 m/s2a = \frac{F_{\text{net}}}{m} = \frac{48.4}{20} \approx 2.42 \, \text{m/s}^2a=mFnet​​=2048.4​≈2.42m/s2

   d) Effect of rougher surface:
   Increasing the coefficient of friction would increase the frictional force, decreasing the net force and reducing the sled's acceleration.